User victor protsak - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T14:37:23Z http://mathoverflow.net/feeds/user/5740 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/128922/volume-of-gr2-4/129028#129028 Answer by Victor Protsak for Volume of Gr(2,4) Victor Protsak 2013-04-28T20:25:10Z 2013-04-28T20:25:10Z <p>The volume of a Grassmanian can be computed using <em>Wirtinger's theorem</em>: </p> <blockquote> <p>The volume of a $p$-dimensional complex submanifold $S$ of a complex Hermitian manifold $(X,\omega)$ is </p> <p>$$\frac{1}{p!}\int_S\omega^p.$$</p> </blockquote> <p>If $X=\mathbb{CP}^N$ the integral is equal to the <em>degree</em> of $S$ times the volume of $X$. Thus up to normalization factors, the volume of the Grassmanian $Gr(k,n)$ is its degree in the Plücker embedding $$Gr(k,n)\subset \mathbb{CP}^N, N=\binom{n}{k}-1.$$</p> http://mathoverflow.net/questions/128212/special-primes-with-p4p1/128283#128283 Answer by Victor Protsak for special primes with p'=4p+1 Victor Protsak 2013-04-21T20:39:14Z 2013-04-21T20:39:14Z <p>Sage (http://www.sagemath.org) has an object type "generator". If you execute </p> <pre> quad=([p,4*p+1] for p in Primes() if is_prime(4*p+1)) </pre> <p>quad.next() will generate consecutive pairs of primes of the form [$p$,$4p+1$]. Here how the list of the first 10 such pairs is generated:</p> <pre> for i in range(0,10): quad.next() [3, 13] [7, 29] [13, 53] [37, 149] [43, 173] [67, 269] [73, 293] [79, 317] [97, 389] [127, 509] </pre> http://mathoverflow.net/questions/121715/invariant-polynomials-for-a-product-of-algebraic-groups/121760#121760 Answer by Victor Protsak for Invariant polynomials for a product of algebraic groups Victor Protsak 2013-02-14T01:18:31Z 2013-02-15T23:07:03Z <p>There is a fruitful way of thinking about the algebra $\mathbb{C}[M_{n,m}]^{{O_n}\times O_m}$ that originates from the $(GL_n, GL_m)$-duality. Namely, </p> <blockquote> <p>$$\mathbb{C}[M_{n,m}]=\bigoplus_{\lambda}V_n^{\lambda}\otimes V_m^{\lambda},$$</p> </blockquote> <p>where the sum is over all partitions $\lambda$ with at most $\min(n,m)$ parts and $V_n^\lambda$ is the polynomial representation of $GL_n$ with highest weight $\lambda=(\lambda_1,\lambda_2,\ldots)$ with zeros appended at the end to make it length $n,$ and similarly for $m.$ Now, since $O_n$ is a spherical subgroup of $GL_n$, the space of $O_n$-invariants $V_n^{\lambda}$ is at most one-dimensional; it is non-zero precisely when $\lambda$ is <em>even</em> (i.e. all parts are even). A good source for these results is Roger Howe's <em>Schur Lectures</em>.</p> <p>It follows that as a vector space, </p> <blockquote> <p>$$\mathbb{C}[M_{n,m}]^{O_n\times O_m}=\bigoplus_{\lambda}(V_n^{\lambda})^{O_n}\otimes (V_m^{\lambda})^{O_m},$$</p> </blockquote> <p>and now the sum is over even partitions with the same restriction as before, and every summand is one-dimensional and is spanned by an explicitly described polynomial on $M_{n,m}$. Using standard techniques (described in the Schur lectures), this description can be amplified to yield the algebra structure as well. Namely, the algebra is graded by an affine semigroup of rank $\min(n,m)$ and is freely generated by explicit elements in degrees $2k$ for $1\leq k\leq \min(n,m).$ With a bit of extra work, one can also handle the more general case of $SO_n\times SO_m$ invariants.</p> <hr> <p>Let me connect this description of the invariants with Abdelmalek's description following the path that Neil originally had in mind. Assume for concreteness that $n\geq m.$ The First Fundamental Theorem for $O_n$ in geometric form states that the map</p> <p>$$q:M_{n,m}\to Sym_{m,m} \qquad X\mapsto X^{T}X$$ is the geometric quotient, i.e. it is $O_n$-invariant and gives rise to the isomorphism</p> <p>$$\mathbb{C}[M_{n,m}]^{O_n}\simeq \mathbb{C}[Sym_{m,m}].$$</p> <p>(The restriction $n\geq m$ assures that the image of $q$, which consists of symmetric matrices of rank at most $\min(n,m),$ is all of $Sym_{m,m}.$)</p> <p>It is also clearly $GL_m$-equivariant, where $GL_m$ acts on the symmetric matrices of order $m$ by </p> <p>$$Y\to g^T Y g\qquad Y\in Sym_{m,m}, g\in GL_m,$$</p> <p>and, <em>a fortiori</em>, $O_m$-invariant. Now pass to the $O_m$-invariants! Under this identification, </p> <blockquote> <p>$$\mathbb{C}[M_{n,m}]^{O_n\times O_m} \simeq \mathbb{C}[Sym_{m,m}]^{O_m}.$$</p> </blockquote> <p>Thus the problem is reduced to determination of the orthogonal invariants of a symmetric matrix $Y$. It is well known that this invariant ring is freely generated by the coefficients of the characteristic polynomial of $Y.$ The coefficient of $\lambda^{m-k}$ has degree $k$ in the entries of $Y$ and, up to a sign, it is equal to the sum of the principal minors of $Y=X^T X$ of order $k, 1\leq k\leq m.$ Thus the fundamental invariants have degrees $2k, 1\leq k\leq m,$ in the entries of $X.$ Additionally, it can be verified that the $k$th fundamental invariant corresponds to the summand with $\lambda=(1^k)$ in the direct decomposition of $\mathbb{C}[M_{n,m}]^{O_n\times O_m}$ coming from the $(GL_n,GL_m)$-duality (i.e. the second highlighted decomposition above).</p> http://mathoverflow.net/questions/121903/how-do-minimal-polynomials-relate/121952#121952 Answer by Victor Protsak for How do minimal polynomials relate? Victor Protsak 2013-02-15T22:27:42Z 2013-02-15T22:27:42Z <p>The general notion of minimal polynomial, as Roy Smith indicated in the comments, works whenever you have an algebra $A$ over a field $K$ and an element $a\in A.$ Namely, consider the unique algebra homomorphism $\pi:k[t]\to A$ that sends $a$ to $t.$* Since $k[t]$ is a principal ideal domain, its kernel $\ker\pi$ is $(P),$ where $P$ may be chosen to be a monic polynomial of degree $d\geq 0$, or 0. In the former case, $f$ is the minimal polynomial of $a$ and in the latter case, $a$ generates a polynomial algebra inside $A.$ The canonical object which perhaps deserves the name "minimal polynomial" is the ideal $(P)$. When $A$ is a general commutative ring, same reasoning leads to the ideal $\ker\pi$, but there is no reason to expect that it is principal.</p> <hr> <p>(*) The idea of "prolonging" the element $a$ to the homomorphism $\pi$ is an algebraic version of the "functional calculus". Thus expressions $q(a)$ for various polynomials $q\in K[t]$ make sense: they are simply $\pi(q).$ However, the functional calculus used in analysis and operator algebra theory involves much larger functional spaces than just polynomials.</p> <hr> <p>Another useful interpretation of the minimal polynomial is through the poles of the formal resolvent of $a.$ For any commutative ring $A$ (not necessarily an algebra over a field), consider </p> <p>$$(\lambda-a)^{-1}=\sum_{n=0}^{\infty}\lambda^{-n-1}a^n \in A((\lambda^{-1})).$$ </p> <p><em>A priori</em>, this is just a Laurent formal power series in $\lambda^{-1}$; however, </p> <blockquote> <p>When $a$ has a minimal polynomial $P(t)$, it is in fact a rational function of $\lambda$ with the denominator (in lowest terms) $P(\lambda),$ and conversely. </p> </blockquote> <p>Indeed,</p> <blockquote> <blockquote> <p>$P(a)(\lambda-a)^{-1}=P(\lambda)(\lambda-a)^{-1}$ modulo polynomials in $\lambda.{\ } \square$</p> </blockquote> </blockquote> <p>See my answer <a href="http://mathoverflow.net/questions/31595/how-would-you-solve-this-tantalizing-halmos-problem/31790#31790" rel="nofollow">http://mathoverflow.net/questions/31595/how-would-you-solve-this-tantalizing-halmos-problem/31790#31790</a> for related ideas.</p> http://mathoverflow.net/questions/118661/good-book-on-representation-theory-of-gln/118677#118677 Answer by Victor Protsak for Good book on representation theory of GL(n) Victor Protsak 2013-01-11T22:14:14Z 2013-01-11T22:14:14Z <p>It's easier to mention the keyword "Littlewood-Richardson coefficients" that gives the answer to your question than to come up with the best possible source explaining it. If you are <em>only</em> interested in the answer, I would suggest to read books on combinatorics, such as "Symmetric group" by Sagan, rather than trying to learn all the background from the representation theory justifying it. </p> http://mathoverflow.net/questions/118090/copies-of-axb-inside-the-an-part-of-an-iwasawa-decomposition/118109#118109 Answer by Victor Protsak for Copies of ax+b inside the AN part of an Iwasawa decomposition? Victor Protsak 2013-01-05T04:48:41Z 2013-01-05T04:48:41Z <p>Here is an elementary argument addressing Q1. The adjoint action of the subalgebra $\mathfrak{a}$ on $\mathfrak{g}$ has the following two key properties:</p> <ol> <li><p>It is diagonalizable (the operators $ad(H)$ for $H\in\mathfrak{a}$ are symmetric with respect to the Killing form);</p></li> <li><p>It stabilizes $\mathfrak{n}$.</p></li> </ol> <p>It follows that either $\mathfrak{a}$ centralizes $\mathfrak{n}$ or there exists an element $H\in\mathfrak{a}$ with a non-zero eigenvalue. In the latter case, rescaling, we get a pair $H\in\mathfrak{a}, X\in\mathfrak{n}$ such that $[H,X]=ad(H)X=X.$ As Yves has already remarked, the former case is never realized since the Lie algebra $\mathfrak{g}$ is semisimple. </p> http://mathoverflow.net/questions/106114/lie-subgroups-of-so2son/106146#106146 Answer by Victor Protsak for Lie Subgroups of SO(2)×So(n) Victor Protsak 2012-09-02T00:44:56Z 2012-09-02T02:30:12Z <p>It is not clear to me what kind of answer is expected. Generally speaking, subgroups of Lie groups can be classified by Lie correspondence combined with combinatorial analysis resulting from structure theory of semisimple Lie groups. Below I address two particular cases.</p> <p>If $K$ is a compact connected <em>semisimple</em> subgroup of $SO(2)\times SO(n)$ then its projection onto the first factor is trivial and the question is reduced to the $SO(n)$ case. (Closed) Lie subgroups of $SO(n)$ are precisely (compact) Lie groups with a faithful $n$-dimensional real orthogonal representation, so there are quite a few of them (the maximal connected ones were classified long time ago by Dynkin). If you need a complete description for small values of $n$, the Atlas of Lie groups is very handy. </p> <p>In the other extreme case where $K=SO(2)$ you are, in effect, asking about the maps </p> <p>$$f: SO(2)\to SO(2)\times SO(n).$$</p> <p>They can be classified by passing to the Lie algebras. More precisely, the differential of $f$ is a linear map $so(2)\to so(2)\oplus so(n).$ Identifying $so(2)$ with $\mathbb{R}$ and $so(n)$ with the skew-symmetric matrices, it may be viewed as a pair $(d,A),$ where $d$ is an integer and $A$ is a skew-symmetric matrix whose eigenvalues are integral multiples of $i.$ Explicitly, </p> <p>$$f:R(\varphi)\mapsto (R(d\varphi), \exp(\varphi A)),$$</p> <p>where </p> <p>$$R(\varphi)=\begin{bmatrix} \phantom{-}\cos(\varphi) &amp; \sin(\varphi) \cr -\sin(\varphi) &amp; \cos(\varphi) \end{bmatrix}$$</p> <p>is the counterclockwise rotation by $\varphi$ and $\exp$ is the matrix exponential function.</p> <p>The maps $f$ and $f'$ associated with non-zero pairs $(d,A)$ and $(d',A')$ have the same image if and only if the pairs are proportional. The case $d=0$ corresponds to an $SO(2)$ subgroup of the second factor $SO(n).$ In the case $d=1$, the subgroup $f(K)$ is the graph of a map $SO(2)\to SO(n).$</p> <p>Note that for the original question about subgroups of $SO(2,n)$ one must impose further equivalences in the case $n=2$, because different subgroups of $SO(2)\times SO(2)$ can be conjugate in $SO(2,2)$. </p> http://mathoverflow.net/questions/37610/demonstrating-that-rigour-is-important/37776#37776 Answer by Victor Protsak for Demonstrating that rigour is important Victor Protsak 2010-09-05T04:21:24Z 2012-05-04T01:01:14Z <p>I would like to preface this long answer by a few philosophical remarks. As noted in the original posting, proofs play multiple roles in mathematics: for example, they assure that certain results are correct and give insight into the problem. </p> <blockquote> <p>A related aspect is that in the course of proving an intuitively obvious statement, it is often necessary to create theoretical framework, i.e. definitions that formalize the situation and new tools that address the question, which may lead to vast generalizations in the course of the proof itself or in the subsequent development of the subject; often it is the proof, not the statement itself, that generalizes, hence it becomes valuable to know multiple proofs of the same theorem that are based on different ideas. The greatest insight is gained by the proofs that subtly modify the original statement that turned out to be wrong or incomplete. Sometimes, the whole subject may spring forth from a proof of a key result, which is especially true for proofs of impossibility statements. </p> </blockquote> <p>Most examples below, chosen among different fields and featuring general interest results, illustrate this thesis.</p> <ol> <li><p><b>Differential geometry</b> </p> <p>a. It had been known since the ancient times that it was impossible to create a perfect (i.e. undistorted) map of the Earth. The first proof was given by Gauss and relies on the notion of intrinsic curvature introduced by Gauss especially for this purpose. Although Gauss's proof of <a href="http://en.wikipedia.org/wiki/Theorema_Egregium" rel="nofollow">Theorema Egregium</a> was complicated, the tools he used became standard in the differential geometry of surfaces.</p> <p>b. Isoperimetric property of the circle has been known in some form for over two millenia. Part of the motivation for Euler's and Lagrange's work on variational calculus came from the isoperimetric problem. Jakob Steiner devised several different synthetic proofs that contributed technical tools (Steiner symmetrization, the role of convexity), even though they didn't settle the question because they relied on the existence of the absolutely minimizing shape. Steiner's assumption led Weierstrass to consider the general question of existence of solutions to variational problems (later taken up by Hilbert, as mentioned below) and to give the first rigorous proof. Further proofs gained new insight into the isoperimetric problem and its generalizations: for example, Hurwitz's two proofs using Fourier series exploited abelian symmetries of closed curves; the proof by Santaló using integral geometry established more general <a href="http://en.wikipedia.org/wiki/Bonnesen%27s_inequality" rel="nofollow">Bonnesen inequality</a>; E.Schmidt's 1939 proof works in $n$ dimensions. Full solution of related lattice packing problems led to such important techniques as Dirichlet domains and Voronoi cells and the geometry of numbers.</p></li> <li><p><b>Algebra</b> </p> <p>a. For more than two and a half centuries since Cardano's <em>Ars Magna</em>, no one was able to devise a formula expressing the roots of a general quintic equation in radicals. The <a href="http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem" rel="nofollow">Abel–Ruffini theorem</a> and Galois theory not only proved the impossibility of such a formula and provided an explanation for the success and failure of earlier methods (cf <a href="http://en.wikipedia.org/wiki/Lagrange_resolvents#Lagrange_resolvents" rel="nofollow">Lagrange resolvents</a> and <a href="http://en.wikipedia.org/wiki/Casus_irreducibilis" rel="nofollow"><em>casus irreducibilis</em></a>), but, more significantly, put the notion of <em>group</em> on the mathematical map.</p> <p>b. Systems of linear equations were considered already by Leibniz. Cramer's rule gave the formula for a solution in the $n\times n$ case and Gauss developed a method for obtaining the solutions, which yields the least square solution in the underdetermined case. But none of this work yielded a criterion for the existence of a solution. Euler, Laplace, Cauchy, and Jacobi all considered the problem of diagonalization of quadratic forms (the <a href="http://en.wikipedia.org/wiki/Principal_axis_theorem" rel="nofollow">principal axis theorem</a>). However, the work prior to 1850 was incomplete because it required genericity assumptions (in particular, the arguments of Jacobi <em>et al</em> didn't handle singular matrices or forms. Proofs that encompass <em>all</em> linear systems, matrices and bilinear/quadratic forms were devised by Sylvester, Kronecker, Frobenius, Weierstrass, Jordan, and Capelli as part of the program of classifying matrices and bilinear forms up to equivalence. Thus we got the notion of rank of a matrix, minimal polynomial, Jordan normal form, and the theory of elementary divisors that all became cornerstones of linear algebra. </p></li> <li><p><b>Topology</b> </p> <p>a. Attempts to rigorously prove the Euler formula $V-E+F=2$ led to the discovery of non-orientable surfaces by Möbius and Listing. </p> <p>b. Brouwer's proof of the <a href="http://en.wikipedia.org/wiki/Jordan_curve_theorem" rel="nofollow">Jordan curve theorem</a> and of its generalization to higher dimensions was a major development in algebraic topology. Although the theorem is intuitively obvious, it is also very delicate, because various plausible sounding related statements are actually wrong, as demonstrated by the <a href="http://en.wikipedia.org/wiki/Lakes_of_Wada" rel="nofollow">Lakes of Wada</a> and the <a href="http://en.wikipedia.org/wiki/Alexander_horned_sphere" rel="nofollow">Alexander horned sphere</a>.</p></li> <li><p><b>Analysis</b> The work on existense, uniqueness, and stability of solutions of ordinary differential equations and well-posedness of initial and boundary value problems for partial differential equations gave rise to tremendous insights into theoretical, numerical, and applied aspects. Instead of imagining a single transition from 99% ("obvious") to 100% ("rigorous") confidence level, it would be more helpful to think of a series of progressive sharpenings of statements that become natural or plausible after the last round of work.</p> <p>a. Picard's proof of the existence and uniqueness theorem for a first order ODE with Lipschitz right hand side, Peano's proof of the existence for continuous right hand side (uniqueness may fail), and Lyapunov's proof of stability introduced key methods and technical assumptions (contractible mapping principle, compactness in function spaces, Lipschitz condition, Lyapunov functions and characteristic exponents). </p> <p>b. Hilbert's proof of the <a href="http://eom.springer.de/D/d032900.htm" rel="nofollow">Dirichlet principle</a> for elliptic boundary value problems and his work on the eigenvalue problems and integral equations form the foundation for linear functional analysis.</p> <p>c. The Cauchy problem for hyperbolic linear partial differential equations was investigated by a whole constellation of mathematicians, including Cauchy, Kowalevski, Hadamard, Petrovsky, L.Schwartz, Leray, Malgrange, Sobolev, Hörmander. The "easy" case of analytic coefficients is addressed by the Cauchy–Kowalevski theorem. The concepts and methods developed in the course of the proof in more general cases, such as the characteristic variety, well-posed problem, weak solution, Petrovsky lacuna, Sobolev space, hypoelliptic operator, pseudodifferential operator, span a large part of the theory of partial differential equations. </p></li> <li><p><b> Dynamical systems </b></p> <p>Universality for one-parameter families of unimodal continuous self-maps of an interval was experimentally discovered by Feigenbaum and, independently, by Coullet and Tresser in the late 1970s. It states that the ratio between the lengths of intervals in the parameter space between successive period-doubling bifurcations tends to a limiting value $\delta\approx 4.669201$ that is independent of the family. This could be explained by the existence of a nonlinear renormalization operator $\mathcal{R}$ in the space of all maps with a unique fixed point $g$ and the property that all but one eigenvalues of its linearization at $g$ belong to the open unit disk and the exceptional eigenvalue is $\delta$ and corresponds to the period-doubling transformation. Later, computer-assisted proofs of this assertion were given, so while Feigebaum universality had initially appeared mysterious, by the late 1980s it moved into the "99% true" category.</p> <p>The full proof of universality for quadratic-like maps by Lyubich (<a href="http://www.ams.org/mathscinet-getitem?mr=1689333" rel="nofollow">MR</a>) followed this strategy, but it also required very elaborate ideas and techniques from complex dynamics due to a number of people (Douady&ndash;Hubbard, Sullivan, McMullen) and yielded hitherto unknown information about the combinatorics of non-chaotic quadratic maps of the interval and the local structure of the Mandelbrot set. </p></li> <li><p><b>Number theory</b></p> <p>Agrawal, Kayal, and Saxena proved that PRIMES is in P, i.e. primality testing can be done deterministically in polynomial time. While the result had been widely expected, their work was striking in at least two respects: it used very elementary tools, such as variations of Fermat's little theorem, and it was carried out by a computer science professor and two undergraduate students. The sociological effect of the proof may have been even greater than its numerous consequences for computational number theory. </p></li> </ol> http://mathoverflow.net/questions/93454/is-there-a-generalization-of-schur-weyl-duality-and-plethysm-for-direct-product/93918#93918 Answer by Victor Protsak for Is there a generalization of Schur - Weyl duality and plethysm for direct product of special unitary groups? Victor Protsak 2012-04-12T23:28:15Z 2012-04-12T23:28:15Z <p>First, a terminological nitpick: the Schur-Weyl duality deals with the <em>unitary</em> group $U(N)$ (in the compact formulation) or the general linear group $GL(N)$ (in the algebraic group version) acting on $\mathcal{H}^{\otimes m}$, where $\mathcal{H}=\mathbb{C}^N.$ The duality states that the isotypic components under the $U(N)$ action are irreducible $S_m$-modules. Explicitly,</p> <p>$$\mathcal{H}^{\otimes m}\simeq \bigoplus_{\lambda}\rho_{U(N)}^\lambda\otimes\rho_{S_m}^{\lambda},$$</p> <p>where $\lambda$ runs over Young diagrams with $m$ boxes and at most $\operatorname{min}(N,m)$ rows. Observe that for $m\geq 3$ and $N\geq 2$ this decomposition is <em>not</em> multiplicity-free as a $U(N)$-module. However, for $N\geq 2$ the restriction to $SU(N)$ does not introduce new multiplicities: for distinct $\lambda,\mu$ as above, the representations $\rho_{SU(N)}^\lambda$ and $\rho_{SU(N)}^\mu$ are non-isomorphic .</p> <p>Now for the present question. Consider the $K$-equivariant isomorphism $$\mathcal{H}^{\otimes m}\simeq \mathcal{H}_1^{\otimes m} \otimes\ldots\otimes\mathcal{H}_s^{\otimes m}.$$ Then each factor $SU(N_i)$ of $K$ acts on its own $m$th tensor power space $\mathcal{H}_i^{\otimes m}.$ It is a standard fact in representation theory that the irreducible representations of $K=SU(N_1)\times\ldots\times SU(N_s)$ have the form $V_1\otimes\ldots\otimes V_s,$ where $V_i$ is an irreducible representation of $SU(N_i)$ determined uniquely up to isomorphism. Hence the question is reduced to the case $s=1.$ Explicitly, $$\mathcal{H}^{\otimes m}\simeq \bigoplus_{\lambda_1,\ldots,\lambda_s}\rho_{SU(N_1)}^{\lambda_1}\otimes\ldots\otimes \rho_{SU(N_s)}^{\lambda_s}\otimes\rho_{S_m}^{\lambda_1}\otimes\ldots\otimes\rho_{S_m}^{\lambda_s}.$$</p> <p>Different $\lambda_i$'s with $m$ boxes are independently chosen, each subject to its own restriction on the number of rows. This is not multiplicity-free unless $m\leq 2$ or all $N_i$ are equal to 1. </p> http://mathoverflow.net/questions/91190/can-the-infinite-von-dyck-groups-be-subgroups-of-sun/91214#91214 Answer by Victor Protsak for Can the infinite von Dyck groups be subgroups of $SU(n)$? Victor Protsak 2012-03-14T19:57:45Z 2012-03-14T20:22:03Z <p>Yes, such embeddings of von Dyck groups (and more generally, of discrete isometry groups of hyperbolic space) can be constructed with the standard machinery from algebraic groups and lattices. Here is a sketch of the construction, but for the sake of simplicity, I'll describe the orthogonal case and only describe the case of the quadratic field. The whole construction, with attributions, is explained well in Lubotzky's book (see also Borel-Wallach). You can treat it as a black box that allows you to embed a subgroup of a group of orthogonal matrices with signature $(n,1)$ and entries in a quadratic field $K$ into the real orthogonal group $O(n+1).$ </p> <p>Let $q$ be the diagonal quadratic form $\sum_{i=1}^{n+1} a_i x_i^2,$ where $a_i$ are non-zero elements of a real quadratic field $K$, and $G=O(q)$ be the isometry group of $q$ viewed as an algebraic group. Denote by $\sigma$ the non-trivial Galois automorphism of the field $K$ and assume that $a_i$ is totally positive for $1\leq i\leq n,$ but $a_{n+1}$ is not. The group $G(K)$ of $K$-points of $G$ embeds into $O(n)\times O(n,1).$ In a plain language, $g\in G(K)$ is a matrix with entries in $K$ that preserves the form $q,$ and it is mapped into the pair of real matrices $(g,g^{\sigma}),$ where $g$ preserves a positively definite quadratic form $q$ and $g^{\sigma}$ preserves the quadratic form $q^{\sigma}$ of signature $(n,1).$ More precisely, by the restriction of scalars from $K$ to $\mathbb{Q}$, $G$ embeds into a product of orthogonal groups over $\mathbb{Q}$ with real forms $O(n+1)$ and $O(n,1),$ and this embedding leads to an embedding $f$ of $G(K)$ into the product of real groups $O(n+1)\times O(n,1).$ Moreover, the composition of $f$ with the projection onto the first factor is an embedding $f'$ of $G(K)$ into $O(n+1).$ </p> <p>Although von Dyck groups are not necessarily arithmetic, they are defined over $\bar{\mathbb{Q}},$ and some of them over quadratic fields. Thus you can embed them into $SO(3),$ although, of course, the image is dense.</p> <p>A standard example is $q=\sum_{i=1}^n x_i^2 +\sqrt{5}x_{n+1}^2,$ with $K=\mathbb{Q}(\sqrt{5}).$ The conjugate form $q^{\sigma}= \sum_{i=1}^n x_i^2 -\sqrt{5}x_{n+1}^2,$ and it's clear that $q$ is positively definite and $q^\sigma$ has signature $(n,1).$ I think, although I haven't checked the details, that the case $n=2$ takes cares of the von Dyck group $(2,5,5).$</p> http://mathoverflow.net/questions/85560/are-nilpotent-orbits-degenerations-of-semi-simple-orbits/85690#85690 Answer by Victor Protsak for Are nilpotent orbits degenerations of semi-simple orbits ? Victor Protsak 2012-01-14T20:51:33Z 2012-01-15T03:18:31Z <p>The answer is often "yes". Here is the sketch of how to obtain a nilpotent orbit as a degeneration of semisimple orbits in the $GL_n$ case. Let $d$ be a partition of $n$ with $k$ parts and $\overline{\mathcal{O}}_{d'}$ be the closure of the conjugacy class of nilpotent $n\times n$ matrices with Jordan blocks sizes given by the dual partition $d'.$ Denote by $\mathcal{O}_d(t_1,\ldots, t_k)$ the conjugacy class of the block diagonal matrix with scalar diagonal blocks $t_i I_{d_i}.$ Then </p> <p>$$\lim_{t\to 0}\ \mathcal{O}_d(t_1,\ldots, t_k)=\overline{\mathcal{O}}_{d'}.$$</p> <p>This is manifested on the level of defining equations using Oshima's approach from </p> <blockquote> <p>A quantization of conjugacy classes of matrices. Adv. Math. 196 (2005), no. 1, 124–146. </p> </blockquote> <p>For a general $\mathfrak{g},$ this amounts to the induction of (zero-dimensional) orbits and to the correspondence between semisimple and regular orbits. In particular, every Richardson nilpotent orbit can be obtained as a degeneration in the same way. However, the defining equations are not known to the same degree of explicitness. </p> <p>On the other hand, if $\mathfrak{g}$ is simple of type other than "A" then the minimal nilpotent orbit is <em>rigid</em>, meaning that it cannot be deformed within the family of adjoint orbits. Existence of rigid orbits makes quantization of orbits a non-trivial task, since a very natural prescription for quantization of semisimple orbits needs to be supplemented by ad hoc quantizations of rigid orbits (several papers of Joseph addressed this question). Rigid orbits have been completely classified: if my memory serves, the answer is in Collingwood-McGovern. </p> http://mathoverflow.net/questions/85664/structure-of-s-mathfrakg-otimes-s-mathfrakgg-for-semisimple-mathfr/85686#85686 Answer by Victor Protsak for Structure of $[S(\mathfrak{g})\otimes S(\mathfrak{g})]^G$ for semisimple $\mathfrak{g}$ Victor Protsak 2012-01-14T20:18:07Z 2012-01-14T20:18:07Z <p>When $G=GL(n),$ this is the "invariants of matrices", as Alexander Chervov has pointed out. The full description by generators and relations is only known for small $n.$ So it's not "textbook material" in the same sense as the description of $S(\mathfrak{g})^G.$ For a general $\mathfrak{g},$ there are papers of Bezrukavnikov-Ginzburg and Kostant, among others, but it's Saturday and cold here as well....</p> http://mathoverflow.net/questions/32099/what-is-lagrange-inversion-good-for/32261#32261 Answer by Victor Protsak for What is Lagrange Inversion good for? Victor Protsak 2010-07-17T04:53:58Z 2011-11-30T17:42:43Z <p>You can use Lagrange inversion to explicitly solve </p> <p>$$x^5-x-a=0\qquad (*)$$ </p> <p>(yes, a <em>fifth</em> degree equation, <em>gasp</em>). More precisely, it yields an infinite series expansion</p> <p>$$x=-\sum_{k\geq 0}\binom{5k}{k}\frac{a^{4k+1}}{4k+1}$$</p> <p>for the root of $(*)$ which is $0$ at $a=0.$ Although this isn't combinatorics, I'd gladly devote a class in any subject I teach to be able to derive it, because by Bring&ndash;Jerrard, any quintic equation can be reduced to this form, and you get a solution of something that many people believe, albeit for differing reasons, to be unsolvable.</p> http://mathoverflow.net/questions/51531/theorems-that-are-obvious-but-hard-to-prove/73799#73799 Answer by Victor Protsak for Theorems that are 'obvious' but hard to prove Victor Protsak 2011-08-26T20:01:58Z 2011-08-26T20:01:58Z <p>The <a href="http://en.wikipedia.org/wiki/Carpenter%27s_rule_problem" rel="nofollow">Carpenter's rule</a>: a planar linkage can be straightened without the links running into each other. Although the statement had initially seemed obvious, its truth or falsity was a matter of debate among the experts for several years until Bob Connelly, Eric Demaine, and Günter Rote finally proved it. (The analogous statement in 3 dimensions is actually false.)</p> http://mathoverflow.net/questions/68667/is-there-an-analog-of-clifford-theorem-in-the-setting-of-lie-algebras/68698#68698 Answer by Victor Protsak for Is there an analog of Clifford Theorem in the setting of Lie algebras? Victor Protsak 2011-06-24T05:58:31Z 2011-06-24T05:58:31Z <p>If the module $M$ is finite-dimensional then the answer seems to be affirmative. </p> <p>Let $Soc_I(M)$ be the maximal semisimple $I$-submodule of $M,$ which is non-zero by the finite-dimensio-nality of $M.$ It is easy to check that it's also $L$-invariant (thinking in Lie groups, this follows from the formula $ngv=g(g^{-1}ng)v, g\in G, n\in N$), hence $Soc_I(M)$ coincides with $M.$ This precisely means that $M$ is completely reducible as an $I$-module.</p> http://mathoverflow.net/questions/68644/what-is-gelfand-tsetlin-basis-for-an-irreducible-representation-of-sln/68697#68697 Answer by Victor Protsak for What is Gelfand-Tsetlin basis for an irreducible representation of sl(n)? Victor Protsak 2011-06-24T05:23:50Z 2011-06-24T05:32:59Z <p>It's more common to talk about the GT basis of $\mathfrak{gl}_n$-modules. In the case of the adjoint representation, the first row of the GT scheme is $(1,0,\ldots,0,-1)$ and each subsequent row satisfies the interlacing condition, which implies that the left (resp, right) edge consists of a string of $1$s (resp $-1$s), followed by a string of $0$s (possibly empty), except that the bottom entry can be $1$ (resp $-1$) if all entries on the left (resp right) are $1$s (resp $-1$), and all other entries are zeros. It follows that the whole scheme can be reconstructed from the positions of the lowest $1$ along the left edge of the triangle and the lowest $-1$ along the right edge. They may occupy any of the $n$ possible rows/positions each, except that both cannot occur in the $n$th row, corresponding to the impossibility of the bottom entry being simultaneously $1$ and $-1.$</p> <p>In your special case $n=3$ the diagrams will look like this:</p> <p><code>$$\begin{array}{rrrrr} 1 &amp; &amp; 0 &amp; &amp; -1\\ &amp; 1 &amp; &amp; 0 &amp; \\ &amp; &amp; 1 &amp; &amp; \end{array} \quad$$</code></p> <p>(this is the highest weight; there are 7 more).</p> http://mathoverflow.net/questions/61377/prehomogeneous-vector-spaces/61450#61450 Answer by Victor Protsak for Prehomogeneous vector spaces Victor Protsak 2011-04-12T19:28:22Z 2011-04-12T19:28:22Z <p>Your two representations of $GL_n$ are not isomorphic, because one of them contains $\Lambda_1$ with multiplicity 2 and the other with multiplicity 1, $\Lambda_1^*$ and $\Lambda_1$ being non-isomorphic. More generally, if $V$ and $W$ are finite-dimensional rational representations of $G=GL_n$ with $V$ irreducible then the multiplicity of $V$ in $W,$ $$\dim\operatorname{Hom}_{G}(V,W),$$ is an invariant of $W$ modulo isomorphism. </p> <p>By the way, the second space is not prehomogeneous, because the bilinear pairing between $\Lambda_1$ and $\Lambda_1^*$ is a non-constant polynomial $GL_n$-invariant. </p> http://mathoverflow.net/questions/57185/nilpotent-matrices-related-to-lie-algebras-of-special-orthogonal-groups-in-charac/57211#57211 Answer by Victor Protsak for Nilpotent matrices related to Lie algebras of special orthogonal groups in characteristic 0 Victor Protsak 2011-03-03T04:58:59Z 2011-03-03T04:58:59Z <p>Let $\lambda$ be a partition of $n.$ Then there exists a skew-symmetric nilpotent matrix whose Jordan blocks sizes are $\lambda_i$ if and only if every even parts has even multiplicity. This follows easily from the representation theory of $\mathfrak{sl}_2$ and is duly recorded in standard sources, e.g. Collingwood and McGovern. It follows that the maximum "nilpotence index" of a skew-symmetric $n\times n$ matrix is $n$ for odd $n$ and $n-1$ for even $n.$</p> http://mathoverflow.net/questions/56631/matrices-characterizing-pairs-ab-ba/56970#56970 Answer by Victor Protsak for Matrices: characterizing pairs $(AB, BA)$ Victor Protsak 2011-03-01T06:13:37Z 2011-03-01T06:13:37Z <p><em>"And are there any nontrivial mathematical applications of this situation?" </em></p> <p>Yes, this is a very important construction in algebraic geometry and representation theory! </p> <ol> <li><p><b>Algebraic geometry.</b> The papers of Kraft and Procesi used this construction to analyze singularities of the closures of nilpotent orbits in the classical Lie algebras <code>$\mathfrak{gl}_n, \mathfrak{sp}_{2n}, \mathfrak{o}_n.$</code> In particular, they proved that, in the case of <code>$\mathfrak{gl}_n,$</code> these closures are normal varieties. Their proof is based on the relation <code>$$\bar{\mathcal{O}}_{\lambda}=r\circ\ell^{-1}(\bar{\mathcal{O}}'_{\mu}). \qquad (*)$$</code> Here <code>$\bar{\mathcal{O}}_{\lambda}$</code> is the closure of the conjugacy class of nipotent $n\times n$ matrices with partition $\lambda$ and <code>$\bar{\mathcal{O}}'_{\mu}$</code> is the closure of the the conjugacy class of nipotent $m\times m$ matrices with partition $\mu,$ where $\mu_i=\operatorname{max}(\lambda_i-1,0)$; the diagram of $\mu$ is obtained from the diagram of $\lambda$ by removing the first column, so that $n-m=\lambda'_1.$ The maps $r$ and $\ell$ are <code>$$r((A,B))=AB, \quad l((A,B))=BA$$</code> in the notation of the question. The papers of Daskiewicz, Kraskiewicz and Przebinda considered a more general situation: starting with $m$ and $n$ and a partition $\mu$ of $m,$ they proved that the formula (*) holds for a certain partition $\lambda$ of $n$. In other words, the algebraic variety <code>$r\circ\ell^{-1}(\bar{\mathcal{O}}'_{\mu})$</code> is the closure of a <em>single</em> nilpotent orbit, without imposing additional assumptions on $m$ and $n;$ the proof involves careful combinatorial analysis, especially when $m>n.$ </p></li> <li><p><b>Representation theory.</b> Without going into too much detail, this construction emerges in Roger Howe's theory of reductive dual pairs. The nilpotent orbits in question arise as the wave front sets or the associated varieties of representations of two classical groups occurring in the Howe duality correspondence. This is considered and exploited in various papers of J.-S. Li, T. Przebinda, P. Trapa, Nishiyama, Oshiai, and Taniguchi, and my own.</p></li> </ol> http://mathoverflow.net/questions/48597/dual-of-the-lie-bracket/48600#48600 Answer by Victor Protsak for Dual of The Lie Bracket Victor Protsak 2010-12-07T23:30:31Z 2010-12-07T23:30:31Z <p>Yes. The dual of the Lie bracket is the exterior differential that maps 1-forms into 2-forms. See any good textbook on differential geometry.</p> http://mathoverflow.net/questions/38856/jokes-in-the-sense-of-littlewood-examples/38930#38930 Answer by Victor Protsak for Jokes in the sense of Littlewood: examples? Victor Protsak 2010-09-16T05:36:32Z 2010-11-21T19:34:39Z <p>We owe Paul Dirac two excellent mathematical jokes. I have amended them with a few lesser known variations. </p> <p>A. <em>Square root of the Laplacian</em>: we want $\Delta$ to be $D^2$ for some first order differential operator (for example, because it is easier to solve first order partial differential equations than second order PDEs). Writing it out,</p> <p><code>$$\sum_{k=1}^n \frac{\partial^2}{\partial x_k^2}=\left(\sum_{i=1}^n \gamma_i \frac{\partial}{\partial x_i}\right)\left(\sum_{j=1}^n \gamma_j \frac{\partial}{\partial x_j}\right) = \sum_{i,j}\gamma_i\gamma_j \frac{\partial^2}{\partial x_i x_j},$$</code></p> <p>and equating the coefficients, we get that this is indeed true if</p> <p><code>$$D=\sum_{i=1}^n \gamma_i \frac{\partial}{\partial x_i}\quad\text{and}\quad \gamma_i\gamma_j+\gamma_j\gamma_i=\delta_{ij}.$$</code></p> <p>It remains to come up with the right $\gamma_i$'s. Dirac realized how to accomplish it with $4\times 4$ matrices when $n=4$; but a neat follow-up joke is to simply define them to be the elements $\gamma_1,\ldots,\gamma_n$ of</p> <p><code>$$\mathbb{R}\langle\gamma_1,\ldots,\gamma_n\rangle/(\gamma_i\gamma_j+\gamma_j\gamma_i - \delta_{ij}).$$</code></p> <hr> <p>Using symmetry considerations, it is easy to conclude that the commutator of the $n$-dimensional Laplace operator $\Delta$ and the multiplication by $r^2=x_1^2+\cdots+x_n^2$ is equal to $aE+b$, where $$E=x_1\frac{\partial}{\partial x_1}+\cdots+x_n\frac{\partial}{\partial x_n}$$ is the Euler vector field. A boring way to confirm this and to determine the coefficients $a$ and $b$ is to expand $[\Delta,r^2]$ and simplify using the commutation relations between $x$'s and $\partial$'s. A more exciting way is to act on $x_1^\lambda$, where $\lambda$ is a formal variable:</p> <p><code>$$[\Delta,r^2]x_1^{\lambda}=((\lambda+2)(\lambda+1)+2(n-1)-\lambda(\lambda-1))x_1^{\lambda}=(4\lambda+2n)x_1^{\lambda}.$$</code></p> <p>Since $x_1^{\lambda}$ is an eigenvector of the Euler operator $E$ with eigenvalue $\lambda$, we conclude that</p> <p><code>$$[\Delta,r^2]=4E+2n.$$</code></p> <hr> <p>B. <em>Dirac delta function</em>: if we can write </p> <p><code>$$g(x)=\int g(y)\delta(x-y)dy$$</code> </p> <p>then instead of solving an inhomogeneous linear differential equation $Lf=g$ for each $g$, we can solve the equations $Lf=\delta(x-y)$ for each real $y$, where a linear differential operator $L$ acts on the variable $x,$ and combine the answers with different $y$ weighted by $g(y)$. Clearly, there are fewer real numbers than functions, and if $L$ has constant coefficients, using translation invariance the set of right hand sides is further reduced to just one, $\delta(x)$. In this form, the joke goes back to Laplace and Poisson.</p> <hr> <p>What happens if instead of the ordinary geometric series we consider a doubly infinite one? Since </p> <p><code>$$z(\cdots + z^{-n-1} + z^{-n} + \cdots + 1 + \cdots + z^n + \cdots)= \cdots + z^{-n} + z^{-n+1} + \cdots + z + \cdots + z^{n+1} + \cdots,$$</code></p> <p>the expression in the parenthesis is annihilated by the multiplication by $z-1$, hence it is equal to $\delta(z-1)$. Homogenizing, we get</p> <p><code>$$\sum_{n\in\mathbb{Z}}\left(\frac{z}{w}\right)^n=\delta(z-w)$$</code></p> <p>This identity plays an important role in conformal field theory and the theory of vertex operator algebras.</p> <hr> <p>Pushing infinite geometric series in a different direction, </p> <p><code>$$\cdots + z^{-n-1} + z^{-n} + \cdots + 1=-\frac{z}{1-z} \quad\text{and} \quad 1 + z + \cdots + z^n + \cdots = \frac{1}{1-z},$$</code></p> <p>which add up to $1$. This time, the sum of doubly infinite geometric series is zero! Thus the point $0\in\mathbb{Z}$ is the sum of all lattice points on the non-negative half-line and all points on the positive half-line: </p> <p><code>$$0=[\ldots,-2,-1,0] + [0,1,2,\ldots]$$</code></p> <p>A vast generalization is given by Brion's formula for the generating function for the lattice points in a convex lattice polytope $\Delta\subset\mathbb{R}^N$ with vertices $v\in{\mathbb{Z}}^N$ and closed inner vertex cones $C_v\subset\mathbb{R}^N$:</p> <p><code>$$\sum_{P\in \Delta\cap{\mathbb{Z}}^N} z^P = \sum_v\left(\sum_{Q\in C_v\cap{\mathbb{Z}}^N} z^Q\right),$$</code></p> <p>where the inner sums in the right hand side need to be interpreted as rational functions in $z_1,\ldots,z_N$.</p> <hr> <p>Another great joke based on infinite series is the <a href="http://en.wikipedia.org/wiki/Eilenberg_swindle" rel="nofollow">Eilenberg swindle</a>, but I am too exhausted by fighting the math preview to do it justice.</p> http://mathoverflow.net/questions/41493/explicit-isomorphism-between-distributions-and-universal-enveloping-algebra/41496#41496 Answer by Victor Protsak for Explicit isomorphism between distributions and universal enveloping algebra Victor Protsak 2010-10-08T09:40:43Z 2010-10-08T09:40:43Z <p>The difficulty with using "explicit monomials" is that this requires a choice of a basis <code>$\{X_1,\ldots, X_{\text{dim}\mathfrak{g}}\}$</code>. Having said that, the PBW theorem states that each element of $U(\mathfrak{g})$ is uniquely represented as a linear combination of monomials in generators of the form $X_{i_1}\ldots X_{i_n}$ with $i_1\leq\ldots\leq i_n$ and $n\geq 0.$ You may identify the elements of this basis of $\mathfrak{g}$ with left invariant vector fields on $G,$ which form a Lie subalgebra of the associative algebra $D(G)$ of differential operators on $G.$ Each monomial thus becomes an element of $D(G)$ and for $D\in D(G),$ the corresponding distribution is $D\cdot\delta_e,$ where $\delta_e$ is the delta function at the identity of $G.$ The effect of a monomial of degree $n$ on a function can be found by the standard calculation involving "integration by parts" $n$ times.</p> http://mathoverflow.net/questions/41389/harmonic-analysis-on-semisimple-groups-modern-treatment/41480#41480 Answer by Victor Protsak for Harmonic analysis on semisimple groups - modern treatment Victor Protsak 2010-10-08T02:21:53Z 2010-10-08T09:09:29Z <p>The first thing one should keep in mind is that harmonic analysis on <em>semisimple</em> Lie groups is very different from the "abstract harmonic analysis" <em>a la</em> Loomis or Hewitt and Ross dealing with locally compact <em>abelian</em> groups. The semisimple case was developed in the large part by Harish-Chandra and his papers (reprinted in his 4 volume <em>Collected papers</em>), while considerably older than Varadarajan's book, are still a good source of results and inspiration for many of us.</p> <p>For an introduction to the subject, I warmly recommend Howe and Tan's book cited by Jim. It treats from the representation theory point of view the simplest nontrivial case, nonabelian harmonic analysis related to $SL(2,\mathbb{R}).$ The book uses elementary tools, and yet it deals with a wide range of topics. Elements of Harish-Chandra's theory for general reductive Lie groups may be found in Howe's survey article <em>A century of Lie theory</em>. </p> <p>The theory of special functions and harmonic analysis on classical symmetric spaces and reductive symmetric spaces based on representation theoretic approach with different flavor is exposed in the books by Vilenkin, Helgasson, and Heckman and Schlichtkrull, which provide good complementary accounts of this theory.</p> http://mathoverflow.net/questions/41142/geometric-interpretation-of-universal-enveloping-algebras/41236#41236 Answer by Victor Protsak for Geometric interpretation of Universal enveloping algebras Victor Protsak 2010-10-06T06:30:59Z 2010-10-06T06:30:59Z <p><b>Orbit method</b> As Mariano and Ben have already mentioned, $U(\mathfrak{g})$ quantizes coadjoint orbits of $\mathfrak{g}.$ </p> <p>In general, for a real algebraic group $G$ with Lie algebra $\mathfrak{g},$ the following three spaces are closely related:</p> <ol> <li><p>The space of primitive ideals of $U(\mathfrak{g}).$ </p></li> <li><p>The set of isomorphism classes of irreducible unitary representations of $G.$</p></li> <li><p>The set of coadjoint orbits of $G.$</p></li> </ol> <p>The original result is due to Kirillov and states that if $G$ is nilpotent and simply-connected then spaces 2 and 3 are in a natural bijection. Later this was extended in various directions by Auslander and Kostant, Gabriel, Borho and Rentschler, Duflo, and many others. For the case <code>$\mathfrak{g}=\mathfrak{gl}_n,$</code> the Jacobson topology on the space of primitive ideals was determined by Oshima in the Advances Math paper. It is related to the topology on the space consisting of Zariski closures of the conjugacy classes of $n\times n$ matrices, but there are some subtle differences.</p> <hr> <p><b>Relation to flag variety</b> In the case when $G$ is a compact semisimple Lie group, the coadjoint orbit <code>$\mathcal{O}_\lambda=G\cdot\lambda$</code> is equivariantly isomorphic to generalized flag variety $G_{\mathbb{C}}/P_\lambda;$ moreover, this isomorphism is compatible with symplectic structures and line bundles. Here $\lambda$ is an integral weight (after natural identification between $\mathfrak{g}$ and its dual) and when $\lambda$ is a regular integral dominant weight, $P_\lambda=B,$ so that the coadjoint orbit <code>$\mathcal{O}_\lambda$</code> may be identified with the full flag variety <code>$G_{\mathbb{C}}/B$</code> polarized by the line bundle $\mathcal{L}_\lambda.$</p> http://mathoverflow.net/questions/40240/elementary-reference-for-algebraic-groups/40415#40415 Answer by Victor Protsak for Elementary reference for algebraic groups Victor Protsak 2010-09-29T05:11:22Z 2010-09-29T05:11:22Z <p>The following is an emended excerpt from my <a href="http://mathoverflow.net/questions/13/learning-about-lie-groups/29458#29458" rel="nofollow">answer</a> to a related question<sup>1</sup> about books about Lie groups for someone with algebraic geometry background. I might add that Procesi's book ideally fits your goals, since you are also interested in representation theory.</p> <blockquote> <p>For someone with algebraic geometry background, I would heartily recommend Procesi's <em>Lie groups: An approach through invariants and representations.</em> It is masterfully written, with a lot of explicit results, and covers a lot more ground than Fulton and Harris. If you like "theory through exercises" approach then Vinberg and Onishchik, <em>Lie groups and algebraic groups</em> is very good (the Russian title included the word "seminar" that disappeared in translation). </p> <p>If you aren't put off by a bit archaic notation and language, vol 2 of Chevalley's <em>Lie groups</em> is still good.</p> </blockquote> <hr> <p><sup>1</sup>That question is exactly one year old and, according to Anton's <a href="http://meta.mathoverflow.net/discussion/684/happy-birthday-mo/#Item_1" rel="nofollow">MO birthday post</a> on meta, was the second "real" question asked on Mathoverflow.</p> http://mathoverflow.net/questions/40082/why-do-we-teach-calculus-students-the-derivative-as-a-limit/40245#40245 Answer by Victor Protsak for Why do we teach calculus students the derivative as a limit? Victor Protsak 2010-09-28T01:26:39Z 2010-09-28T01:26:39Z <p>I am surprised that no answer has explicitly mentioned the fundamental theorem calculus yet: that is a classic, and important, instance of calculating the derivative using the limit definition. So, for example, the integral sine function</p> <p>$$\int_0^x \frac{\sin t}{t} dt$$ </p> <p>has important applications in signal processing and the cumulative distribution function of the normal distribution $N(a,\sigma^2)$ </p> <p>$$\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^x e^{-\frac{(t-a)^2}{2\sigma^2}}dt$$</p> <p>is the bread and butter of probability and statistics. Both functions are not elementary and their derivatives, while significant, would be impossible to calculate by other means.</p> <p>I also disagree with the comment that piecewise defined functions "are not good at all" for illustrating the definition of the derivative based on limits. In fact, piecewise polynomial functions, in the form of splines, are used in mechanical engineering (e.g. to design the shape of the car body), and provide a neat opportunity to relate conceptual and computational aspects of derivatives. </p> http://mathoverflow.net/questions/39692/are-there-any-uses-for-complex-sine-sin-z/39707#39707 Answer by Victor Protsak for Are there any uses for complex sine? [sin z] Victor Protsak 2010-09-23T07:07:45Z 2010-09-23T07:07:45Z <p>Euler discovered an infinite product expansion for the sine function,</p> <p>$$\sin z=z\prod_{k\geq 1}\left(1-\frac{z^2}{k^2\pi^2}\right)$$ by analogy with the factorization of a polynomial with known zeroes (i.e. roots) into linear terms. In order for this to be true, it is crucial to know <em>all</em> zeros, real as well as complex. As a consequence of the product formula, Euler evaluated zeta function at even integers in terms of Bernoulli numbers. </p> <p>Euler gave an interesting proof of the product formula based on the idea with zeros, where it is shown that the products of $n$ factors approximate $\sin z$ as $n\to\infty.$ I've read some recent papers making this proof entirely rigorous (e.g. cited in Varadarajan's Bull of AMS article). The first step is to write</p> <p>$$\sin z=\frac{1}{2i}(e^{iz}-e^{-iz})= \frac{1}{2i}\lim_{n\to\infty}\left(\left(1+\frac{iz}{n}\right)^n-\left(1-\frac{iz}{n}\right)^n\right).$$</p> <p>A standard proof of the product formula in complex analysis textbooks (due to Weierstrass?) also crucially relies on the fact that $\sin z$ is a complex analytic function, rather than merely real analytic.</p> http://mathoverflow.net/questions/39194/generators-for-the-algebra-of-gln-equivariant-maps-from-m-n-m-n-to-m-n/39291#39291 Answer by Victor Protsak for Generators for the algebra of GL(n)-equivariant maps from M_n + M_n to M_n Victor Protsak 2010-09-19T10:34:49Z 2010-09-19T10:34:49Z <p>The answer is affirmative not only in the case of 2 matrices, but also in the case of any number of matrices; in fact, an analogous statement is true for quiver representations (in characteristic 0).</p> <p>The original question can be restated as follows.</p> <blockquote> <p>Let $P$ be the space of polynomial functions of 2 $n\times n$ matrices, with the adjoint action of $GL_n$ and the ring of invariants $I.$ Consider the space $\text{Hom}_{GL_n}(M_n,P)$ as an $I$-module. Is it true that it is generated by the products of matrices?</p> </blockquote> <p>For the case of any number of generic matrices $A_1,\ldots,A_k,$ Procesi proved that over a field $k$ of characteristic 0, $I$ is spanned by the traces of the products of matrices. Formally, consider words in the free monoid with $k$ generators, substitute the generic matrices, and take a trace. </p> <p>Procesi, C. <em>The invariant theory of n×n matrices</em>. Advances in Math. 19 (1976), no. 3, 306&ndash;381</p> <p>The statement follows by adjoining an extra generic matrix $A_0$ and converting an $M_n$-space into a $GL_n$-invariant forming a product with $A_0$ and taking the trace, then undoing the trace of the term in the trace polynomial from Procesi's theorem containing $A_0.$ </p> <hr> <p>Here is a vast generalization due to Le Bruyn and Procesi. Given a finite quiver $Q$ and a dimension vector $\alpha,$ consider the corresponding representation space $R(Q,\alpha)$ with the action of the algebraic group $GL(\alpha)$ and the space $P$ of polynomial functions on $R.$ (If the quiver consists of a single vertex with $k$ loops and $\alpha=n$ then the representation space is given by $k$ generic $n\times n$ matrices with the simultaneous conjugation action by $GL_n.$) Then, over a field of characteristic 0, the algebra $I$ of polynomial invariants is spanned by the traces of matrix products over oriented cycles in $Q$ and for any pair of vertices $(i,j)$ of $Q,$ the space <code>$\text{Hom}_{GL(\alpha)}(\text{Hom}_k(V_i,V_j),P)$</code> is generated as an $I$-module by the products over oriented paths connecting $i$ with $j.$</p> <p>Lieven Le Bruyn, Claudio Procesi, <em>Semisimple representations of quivers</em>. Trans. Amer. Math. Soc. 317 (1990), no. 2, 585&ndash;598</p> http://mathoverflow.net/questions/39056/book-on-symplectic-geometry/39060#39060 Answer by Victor Protsak for Book on Symplectic Geometry Victor Protsak 2010-09-17T05:44:44Z 2010-09-17T05:59:46Z <p>If you are physically inclined, V.I.Arnold's <em>Mathematical methods of classical mechanics</em> provides a masterful short introduction to symplectic geometry, followed by a wealth of its applications to classical mechanics. The exposition is much more systematic than vol 1 of Landau and Lifschitz and, while mathematically sophisticated, it is also very lucid, demonstrating the interaction between physical ideas and mathematical concepts that support them. (It is also worth mentioning that Arnold was largely responsible for the reawakening of interest to symplectic geometry at the end of 1960s and pioneered the study of symplectic topology. Some of these developments were brand new when the book was first published in 1974 and are briefly discussed in the appendices).</p> <p>In addition to the notes by Cannas da Silva mentioned by Dick Palais, here are further two advanced books covering somewhat different territory:</p> <blockquote> <p>Michèle Audin, <em>Torus actions on symplectic manifolds</em> (2nd edition)<sup><b>A</b></sup></p> <p>Dusa McDuff and Dietmar Salamon, <em>Introduction to symplectic topology</em></p> </blockquote> <hr> <p><sup><b>A</b></sup> In her book, Michèle Audin herself recommends </p> <blockquote> <p>Paulette Libermann and Charles-Michel Marle, <em>Symplectic geometry and analytical mechanics</em></p> </blockquote> <p>as a wonderful introduction to symplectic geometry. </p> http://mathoverflow.net/questions/38943/an-orthogonal-companion-matrix/38989#38989 Answer by Victor Protsak for An orthogonal companion matrix Victor Protsak 2010-09-16T16:22:45Z 2010-09-16T16:22:45Z <p>You can use the <a href="http://en.wikipedia.org/wiki/Cayley_transform" rel="nofollow">Cayley transform</a> and reduce this problem to generating a skew-symmetric matrix with a prescribed characteristic polynomial. For example, this works in the $3\times 3$ case (although when $n$ is odd, $1$ is always an eigenvalue). I have not thought about it thoroughly, but presumably, methods used in Inverse Symmetric Eigenvalue Problem should apply in the skew-symmetric case. </p> http://mathoverflow.net/questions/130960/are-residually-finite-perfect-groups-residually-alternating/130967#130967 Comment by Victor Protsak Victor Protsak 2013-05-17T17:35:59Z 2013-05-17T17:35:59Z By the congruence subgroup property, any finite quotient of $SL_n(\mathbb{Z})$ factors through $SL_n(\mathbb{Z}/N\mathbb{Z})$ for some $N.$ But how do you conclude that any finite simple quotient factors through $SL_n(\mathbb{Z}/p\mathbb{Z})$ for some prime $p$? Also, a nitpick: it is $PSL_n(\mathbb{Z}/p\mathbb{Z})$ that is simple as an abstract group. http://mathoverflow.net/questions/130851/sage-or-magma-implementation-of-nilpotent-orbit-varieties Comment by Victor Protsak Victor Protsak 2013-05-16T21:10:11Z 2013-05-16T21:10:11Z Can you, please, clarify what kind of operations or properties you are interested in? Also, I am not an expert, but Macaulay seems more likely to have something of this kind. http://mathoverflow.net/questions/130522/finding-spherical-representations-of-gln-mathbbc Comment by Victor Protsak Victor Protsak 2013-05-14T06:00:34Z 2013-05-14T06:00:34Z What do you mean by &quot;fix vectors in $\mathbb{C}$? If $\mathbb{C}$ denotes the trivial representation of the group then the answer is also trivial (namely, all of them). Is the group $G$ a complex Lie group (like $GL(n,\mathbb{C})$) or do you simply wish to consider complex representations (not necessarily on $\mathbb{C}$ - perhaps, you can specify whether they are supposed to be finite-dimensional). By the way, $U_n$ is <i>not</i> a complex Lie group, in fact, it is compact, and so it coincides with its own maximal compact subgroup $K.$ http://mathoverflow.net/questions/130527/what-is-the-fundamental-theorem-of-invariant-theory Comment by Victor Protsak Victor Protsak 2013-05-14T05:46:09Z 2013-05-14T05:46:09Z I agree that the question resembles Jabberwocky. What is the relation between representations of $U(N_f)$ and $U(N_c)$? Standard linear algebra operations (direct sums, tensor products, symmetrizations, etc) would transform representations of a group $G$ (here $U(N_f)$) into representations of the <i>same</i> group $G$. http://mathoverflow.net/questions/129044/defining-equations-of-a-flag-variety Comment by Victor Protsak Victor Protsak 2013-04-29T01:03:14Z 2013-04-29T01:03:14Z Although I am not sure that the precise answer is there (and I don't have it close by hand to check), &quot;Combinatorial Commutative Algebra&quot; by Miller and Sturmfels discusses many related questions. http://mathoverflow.net/questions/126705/sh-lie-algebra-cohomology Comment by Victor Protsak Victor Protsak 2013-04-06T23:40:52Z 2013-04-06T23:40:52Z What does &quot;sh&quot; stand for here? http://mathoverflow.net/questions/121366/d-affine-morphisms-and-composition/121510#121510 Comment by Victor Protsak Victor Protsak 2013-02-12T00:29:58Z 2013-02-12T00:29:58Z What do you mean by &quot;Projective spaces and flag varieties don't live in interesting families&quot;? For example, the projectivization $P(W)$ of a rank n vector bundle W over base Y is a family of n-1-dimensional projective spaces that is often not a direct product. http://mathoverflow.net/questions/121539/properties-of-artin-shephard-and-generalized-von-dyck-groups Comment by Victor Protsak Victor Protsak 2013-02-12T00:20:19Z 2013-02-12T00:20:19Z What is the question? Is it the status of Q1-Q3? http://mathoverflow.net/questions/120713/decomposition-of-regular-representation-of-non-compact-lie-group Comment by Victor Protsak Victor Protsak 2013-02-04T23:15:27Z 2013-02-04T23:15:27Z Rather than adding yet another answer, let me just highlight the relevant term &quot;Plancherel measure&quot;. Unitary representations appearing in the decomposition of the regular representation, i.e. those in the support of the Plancherel measure, are known as &quot;tempered representations&quot; and form an important part of the unitary dual. See e.g. Wallach's book. As Paul Garrett explained, it is known explicitly for complex semisimple Lie groups after Gelfand-Naimark's pioneering work and in many other cases, including connected nilpotent groups via Kirillov's orbit method. http://mathoverflow.net/questions/120236/interesting-examples-of-minimal-action-on-torus Comment by Victor Protsak Victor Protsak 2013-01-29T18:28:08Z 2013-01-29T18:28:08Z No: finite groups have finite (and hence closed) orbits. http://mathoverflow.net/questions/118117/can-every-curve-be-written-as-fxgy/118134#118134 Comment by Victor Protsak Victor Protsak 2013-01-05T19:05:32Z 2013-01-05T19:05:32Z Correction: generic curves of genus $g$ depend on $3g-3$ parameters. http://mathoverflow.net/questions/108886/erratum-for-fulton-and-harris/108909#108909 Comment by Victor Protsak Victor Protsak 2012-10-06T01:37:42Z 2012-10-06T01:37:42Z Ryan: it can be placed to Wikibooks that abides by weaker policies. (An even bolder proposal: convince Wikimedia to open Project Errata.) http://mathoverflow.net/questions/106114/lie-subgroups-of-so2son/106146#106146 Comment by Victor Protsak Victor Protsak 2012-09-20T19:49:15Z 2012-09-20T19:49:15Z The first factor is abelian and semisimple groups do not have non-trivial characters. http://mathoverflow.net/questions/107577/are-certain-simple-lie-groups-linear-algebraic-groups Comment by Victor Protsak Victor Protsak 2012-09-19T21:23:36Z 2012-09-19T21:23:36Z You are right, $PGL_2(R)$ is $SO(1,2).$ I guess the reason (connectedness in Zariski vs real topology) was not immediately clear from your first comment. http://mathoverflow.net/questions/107577/are-certain-simple-lie-groups-linear-algebraic-groups Comment by Victor Protsak Victor Protsak 2012-09-19T19:23:35Z 2012-09-19T19:23:35Z Another problem: the connected component of identity in a non-compact orthogonal group $O(p,q)$ has index 4, whereas the algebraic connected component $SO(p,q)$ has index 2.