most recent 30 from http://mathoverflow.net 2013-05-22T14:21:48Z http://mathoverflow.net/feeds/user/4542 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111404/algorithm-to-check-is-representation-irreducible-algorithm-to-decompose-the-re/111419#111419 Andy B 2012-11-04T01:44:50Z 2012-11-04T16:08:23Z

<p>I don't know how to do this exactly, but one could try a probabilistic approach. Say the representation is $V$, and $V$ is defined over a $\mathbb{C}$ (or at least $\mathbb{Q}$). Pick a random non-zero element $v \in V$ and compute the dimension of the space spanned by the orbit $Gv$. If this dimension is $&lt; \dim V$ then $V$ is reducible and (1) is answered. If not, try again. I would guess that if, after several iterations, the subrepresentations are all equal to $V$, then with high probability $V$ is <i>multiplicity free</i> (Edit. This did read "irreducible". As pointed out in the comments by Daniel Litt this is not so!).</p> <p>The worst part about this algorithm (aside from not detecting reducibility) is having to list the elements of $G$. But I guess this is already done according to Alexander's statement of the problem. You could also try to approximate the orbit. Say if $\dim V = 1000$ and $G = S_{12}$, then take some random $2000$ elements of $G$ and apply them $v$, and let this be your approximate orbit.</p> <p>Another bad thing about this algorithm is that it can be numerically unstable.</p>

http://mathoverflow.net/questions/104422/cohomology-and-j Andy B 2012-08-10T17:22:33Z 2012-08-10T19:24:45Z

<p>I have a projective variety $X$ and an open immersion $j : U \to X$.</p> <p>Say I have a sheaf, locally free in my case of interest, $\mathcal{S}$ on $U$. Is there any reasonable relationship between $H^i(X,j_! \mathcal{S})$ and $H^i(U,\mathcal{S})$? What if I add that I know that $H^i(U,\mathcal{S}) = 0$ for $i>0$. I'm hopeful the latter can imply that $H^i(X,j_! \mathcal{S}) = 0 $ for $i>0$.</p>

http://mathoverflow.net/questions/86477/simple-question-in-the-representation-of-sl2-c/86488#86488 Andy B 2012-01-23T21:48:22Z 2012-01-23T21:48:22Z

<p>You're looking at plethysm of $SL_2(\mathbb{C})$-modules. According to a paper of Manivel (An extension of the Cayley-Sylvester formula, 2008) the answer is given by the Cayley-Sylvester formula. In your case it states that the multiplicity of $Sym^e(V)$ in $Sym^n(Sym^3(V))$ is $$ Par(n,3;(3n-e)/2) - Par(n,3;(3n-e)/2 - 1), $$ where $Par(n,k,m)$ is the number of partitions in an $n$-by-$k$ box of size $m$. For example, if $n=3$ and $e=5$ then $Par(3,3;2)=2$ and $Par(3,3;1) = 1$, which agrees with what you have above.</p>

http://mathoverflow.net/questions/85985/symmetric-polynoms-are-hopf-algebra-what-for-one-needs-co-product/86037#86037 Andy B 2012-01-18T21:48:34Z 2012-01-18T21:48:34Z

<p>I'm surprised nobody has mentioned the connection to the Littlewood-Richardson coefficients so far in response to "What is it useful for?". The coproduct $$ \Delta(h_k) = \sum_{i+j = k} h_i \otimes h_j $$ gives rise to the following formula $$ \Delta( s_\lambda ) = \sum_{\mu,\nu} c_{\mu,\nu}^{\lambda} s_\mu \otimes s_\nu $$ Here $c_{\mu,\nu}^{\lambda}$ is the Littlewood-Richardson coefficient and $s_\lambda$ is the Schur function of shape $\lambda$. Hopf algebra techniques have been used to derive "skew" Pieri rules recently in work of Lam, Lauve and Sottile.</p>

http://mathoverflow.net/questions/76497/schur-weyl-duality/76542#76542 Andy B 2011-09-27T18:32:07Z 2011-09-27T18:32:07Z

<p>The multiplicity of a Specht module $Specht(\lambda)$ (Fulton-Harris Ch.4) in $V^{\otimes n}$ is the number semi-standard Young tableaux of shape $\lambda$ and entries in ${1,2,\dots,\dim V}$. In your example (where I've guessed you're assuming $\dim V = 3$) the multiplicity of the Spetch module $Specht((2,1))$ is the number of SSYT of this shape with entries in ${1,2,3}$. This is computed using the hook-content formula, and is 8 (or just write down the 8 tableuax).</p> <p>It is instructive to find a basis of the Schur module of shape $\lambda$ indexed by SSYT of shape $\lambda$. See Fulton's book on Young tableuax for this, or the book Constructive invariant theory by Sturmfels. </p>

http://mathoverflow.net/questions/1887/given-a-spanning-tree-and-an-edge-not-on-the-spanning-tree-how-to-form-a-cycle-b/70252#70252 Andy B 2011-07-13T18:45:44Z 2011-07-13T18:45:44Z

<p>Although this thread is long-dead, there is a very simple way to do this with linear algebra, so I will post an answer.</p> <p>You are interested in a basis of the cycle space of $G$. Orient $G$ in any way. Let $\partial(G)$ be the edge-vertex incidence matrix of $G$. A basis of the cycle space is given by a basis for $\ker \partial(G)$. To get the answer to your question note that if you take any spanning tree $T$ and you consider the edges of $E-T$, there will be a unique element of the cycle space that is $1$ on a given edge of $E-T$ and $0$ on the others. Finding such an element for each edge of $E-T$ will yield a new basis for $\ker \partial(G)$.</p>

http://mathoverflow.net/questions/68263/principal-minors/68267#68267 Andy B 2011-06-20T04:53:58Z 2011-06-20T21:25:20Z

<p>For the incidence matrix $\partial(G)$ of a graph $G$ the answer is easy, as this amounts to knowing how many spanning trees $G$ possess. One can use the matrix-tree theorem to compute the answer by evaluating the determinant of the (reduced) Laplacian matrix $(\partial(G)\partial(G)^T)^{vv}$ of $G$.</p> <p>For a full rank unimodular matrix $M$ (i.e., all maximal minors are $\pm 1$ or zero) with rational entries there is a related "matrix-tree" type theorem: The number of maximal non-zero minors of $M$ is $\det(M M^T)$. This is probably most useful when the matrix handed to you is <em>a priori</em> unimodular, as above. </p> <p>For a matrix $M$ which fails to be unimodular the number $\det(MM^T)$ gives an upper bound. </p>

http://mathoverflow.net/questions/67171/hilbert-mumford-criterion-and-closedness/67197#67197 Andy B 2011-06-08T01:06:33Z 2011-06-08T01:06:33Z

<p>I have a counterexample now, thanks to some notes of Zinovy Reichstein I found. I think the counterexample is paraphrased as follows: Let $V$ be an irreducible representation of $G$ and suppose $x$ does not have a highest weight vector $y$ in its orbit, but $y \in \overline{Gx}$. There will be no way to get to $y$ from $x$ using only semi-simple elements. </p> <p>To be precise (and to take the example from Reichstein) let $G = {\rm SL}_2(\mathbb{C})$ and $V = {\rm{Sym}} ^n \mathbb{C}^2$, $n \geq 2$. If $a$ and $b$ are the standard basis vectors of $\mathbb{C}^2$ then $a^{n-1} b \in V$ has the highest weight vector $a^n$ in its orbit closure, but not its orbit. Now check that there is no one-parameter subgroup $\lambda(t)$ such that $\lambda(t) a^{n-1} b \in G a^n$.</p>

http://mathoverflow.net/questions/67171/hilbert-mumford-criterion-and-closedness Andy B 2011-06-07T18:10:13Z 2011-06-08T01:06:33Z

<p>A version of the Hilbert-Mumford criterion states the following: Let $G$ be a linearly reductive group and $V$ a representation of $G$ over a field $k$ (alg. closed, char. zero). Suppose that $y \in \overline{Gx} - Gx$. Then, there is a one-parameter subgroup $\lambda : k^\times \to G$ such that $$ \lim_{t\to 0} \lambda(t)x \in \overline{Gy}. $$</p> <p>My question is: Is there an example where every one parameter subgroup misses the orbit of $y$? I.e., is there an example where, for every $\lambda: k^\times \to G$ $$ \lim_{t\to 0} \lambda(t)x \in \overline{Gy} \implies \lim_{t\to 0} \lambda(t)x \in \overline{Gy}-Gy? $$ If $G$ is a torus the answer is "no". What if $V$ is replaced by a more general scheme $X$ that is not itself a representation?</p>

http://mathoverflow.net/questions/55436/axxb-and-the-cecioni-frobenius-theorem Andy B 2011-02-14T19:52:42Z 2011-03-09T00:05:51Z

<p>The Frobenius--Cecioni theorem states that if $A$ and $B$ are square matrices with entries in a field $k$ then the dimension of the $k$ vector space of solutions of $$ AX=XB $$ is given by the sum $$ \sum_{i,j} \deg( \gcd(d_i(A),d_j(B))) $$ Here $d_i(\cdot)$ denotes the $i$th invariant factor of its argument. My question: How well known is this theorem in any particular context? (E.g., "I use it all the time and I work in number theory ...", "I've never heard of it and I taught abstract algebra for years ...") Are there any well known applications of this result?</p>

http://mathoverflow.net/questions/1722/free-high-quality-mathematical-writing-online/46592#46592 Andy B 2010-11-19T05:15:39Z 2010-11-19T05:15:39Z

<p>Paul Garrett is quite the author:</p> <p><a href="http://www.math.umn.edu/~garrett/" rel="nofollow">http://www.math.umn.edu/~garrett/</a></p> <p>He has a book on buildings and many vignettes about automorphic forms, L-functions, representation theory, .... He wrote a graduate algebra book while he taught the course, and promptly got it <a href="http://www.amazon.com/Abstract-Algebra-Paul-B-Garrett/dp/1584886897/ref=sr_1_1?ie=UTF8&amp;qid=1290143567&amp;sr=8-1" rel="nofollow">published</a>.</p> <p><a href="http://www.math.umn.edu/~garrett/m/algebra/" rel="nofollow">http://www.math.umn.edu/~garrett/m/algebra/</a></p>

http://mathoverflow.net/questions/44951/is-there-an-account-of-the-algebra-of-highest-weight-tensors/45255#45255 Andy B 2010-11-08T03:17:54Z 2010-11-08T03:17:54Z

<p>The highest weight vectors in the tensor algebra are obtained by applying certain Young symmetrizers to tensor products of standard basis vectors of $V$. For example, a highest weight vector is obtained from $$ e_1 \otimes e_1 \otimes e_2 \otimes e_2 $$ by applying the Young symmetrizer of the tableau of shape $(2,2)$ filled with $1,2,3,4$ from left-to-right top-to-bottom. There is a basis for the space of highest weight vectors indexed by pairs of tableaux $(T,S)$ where $T$ and $S$ have the same shape, row $i$ of $T$ consists of the number $i$ and $S$ is standard. I think this is written up in Fulton's book on Young Tableaux.</p> <p>The product of two highest weight vectors must be "straightened" to be expressed as a integer linear combination of highest weight vectors of this form. This straightening algorithm is well know (in complete generality by Grosshans--Rota--Stein, but this case might be due to De Concini and Procesi, perhaps {way} eariler. You might look in Brian Taylor's MIT thesis to see that such relations form a Groebner basis).</p>

http://mathoverflow.net/questions/39296/matrix-products-under-which-the-determinant-behaves-multiplicatively/39330#39330 Andy B 2010-09-19T19:47:12Z 2010-09-19T19:47:12Z

<p>The determinant of the product of two <i>non square</i> matrices is nicely expressed by the Binet-Cauchy formula: $$ \det(AB) = \sum_I \det A_I \det B_I $$ Here $A$ is $n \times m$ and $B$ is $m \times n$ and the sum ranges over $n$-subsets $I$ of the numbers ${1,2,...,m}$. $A_I$ means "select columns of $A$ indexed by $I$" and $B_I$ means "select the rows of $B$ indexed by $I$". If either $A$ or $B$ has rank less than $n$ than the determinant of $AB$ is, thus, zero.</p> <p>I do not know for certain, but this looks like it has to do with some kind of coproduct?</p>

http://mathoverflow.net/questions/2144/a-single-paper-everyone-should-read/37642#37642 Andy B 2010-09-03T17:01:46Z 2010-09-03T17:01:46Z

<p><a href="http://www-cs-faculty.stanford.edu/~uno/papers/tnn.tex.gz" rel="nofollow">Two notes on notation</a> by Knuth. This paper discusses "Iverson" notation, which is of use to almost all mathematicians, and good notation for Stirling numbers.</p>

http://mathoverflow.net/questions/30981/definition-of-and-intuition-for-regular-subdivisions-of-a-polytope/30997#30997 Andy B 2010-07-08T05:23:04Z 2010-07-08T06:50:33Z

<p>A very good reference for this is the recent book of De Loera, Rambau and Stantos called "Triangulations: Structures for Algorithms and Applications". They have a whole chapter on regular triangulations. I don't know if the book is in print yet. It's possible that careful googling could reveal drafts of the book that are still online.</p> <p>The classic example of a triangulation of a point set that is not regular is the following: Draw two concentric triangles with vertices {1,2,6} and {3,4,5}. The following is a triangulation: {16,26,21, 34,45,43, 36,56, 13,14, 24,25}, where {ij} means draw an edge from i to j.</p> <p>You can see that this is not regular by assuming that {3,4,5} were not lifted above the plane, and then trying to lift vertices, 1,2,6 to get the remaining faces. You will find that you need the heights to satisfy height(1) &lt; height(2) &lt; height(6) &lt; height(1). This is spelled out in complete detail in <a href="http://books.google.com/books?id=J5cVkNIbgXgC&amp;lpg=PP1&amp;dq=sturmfels&amp;pg=PA64#v=onepage&amp;q=regular&amp;f=false" rel="nofollow">Sturmfels' book</a>.</p>

http://mathoverflow.net/questions/22089/enumeration-of-regular-graphs/22109#22109 Andy B 2010-04-21T20:15:11Z 2010-04-21T20:15:11Z

<p>I think the answer is no, but I would consult the following link:</p> <p><a href="http://www.mathe2.uni-bayreuth.de/markus/reggraphs.html" rel="nofollow">http://www.mathe2.uni-bayreuth.de/markus/reggraphs.html</a></p> <p>which contains tables of the sums of the numbers you are interested in. The author is very generous with sharing data that is not posted online.</p>

http://mathoverflow.net/questions/17778/books-you-would-like-to-see-translated-into-english/17814#17814 Andy B 2010-03-11T03:49:48Z 2010-03-11T03:49:48Z

<p>Chirurgie des grassmanniennes by Laurent Lafforgue.</p> <p><a href="http://www.ihes.fr/~lafforgue/math/M02-45.pdf" rel="nofollow">http://www.ihes.fr/~lafforgue/math/M02-45.pdf</a></p>

http://mathoverflow.net/questions/130749/reference-quote-request-all-of-combinatorics-is-the-representation-theory-of-s Andy B 2013-05-15T19:09:00Z 2013-05-15T19:09:00Z

Reminds me of a thing I just read: <a href="http://www.math.ucla.edu/~pak/hidden/papers/Quotes/Combinatorics-quotes.htm" rel="nofollow">math.ucla.edu/~pak/hidden/papers/Quotes/&hellip;</a>

http://mathoverflow.net/questions/127808/a-basis-of-the-symmetric-power-consisting-of-powers Andy B 2013-05-01T19:15:11Z 2013-05-01T19:15:11Z

The best reason I know of that powers of linear forms span $Sym^k(V)$ is that $Sym^k(V)$ is an irreducible $GL(V)$-module. Hence, the orbit of any particular (non-zero) power must span the entire space. After choosing the standard basis for $V$ as $e_1,\dots,e_n$ and viewing $GL(V)$ as square matrices, a highest weight vector for $Sym^k(V)$ is $e_1^k$. Hence, to obtain a basis one only needs to apply lowering operators to this element (which are lower triangular matrices). The proper choice of lowering operators will yield a basis.

http://mathoverflow.net/questions/111404/algorithm-to-check-is-representation-irreducible-algorithm-to-decompose-the-re/111419#111419 Andy B 2012-11-04T16:01:52Z 2012-11-04T16:01:52Z

@daniel litt. good point. in haste i'd thought that by taking a random isomorphic copy of the representation you could get rid of this problem. i guess it makes the problem worse(!), since the rep is now not even visibly reducible. as derek holt points out, you could factorize the characteristic polynomial to see that this is reducible.

http://mathoverflow.net/questions/108419/smith-normalform-of-a-matrix-with-1-outside-the-diagonal Andy B 2012-09-30T19:22:39Z 2012-09-30T19:22:39Z

Also, when all the $a_i = n$ this becomes the computation of the critical group of the the complete graph. The SNF in this case is $(1,n,\dots,n,0)$.

http://mathoverflow.net/questions/108419/smith-normalform-of-a-matrix-with-1-outside-the-diagonal Andy B 2012-09-30T19:20:36Z 2012-09-30T19:20:36Z

My guess is that the answer depends subtly on the gcd of the ai (or worse, subsets of the ai) try various cases where the ai are all distinct primes.

http://mathoverflow.net/questions/104643/tips-for-turning-an-academic-cv-into-a-resume-for-non-academic-jobs Andy B 2012-08-13T21:28:04Z 2012-08-13T21:28:04Z

You should list the programming languages you know. e.g., matlab, c, fortran, mathematica, r, python, ... Human languages (other than English) too.

http://mathoverflow.net/questions/104422/cohomology-and-j/104431#104431 Andy B 2012-08-10T21:24:36Z 2012-08-10T21:24:36Z

Thanks for the example. This is the sort of set-up I've been working with. I guess some of the particulars of my sheaves and spaces will need to come into play.

http://mathoverflow.net/questions/104422/cohomology-and-j/104430#104430 Andy B 2012-08-10T20:12:58Z 2012-08-10T20:12:58Z

I did mean, although I did not say, locally free sheaves of $\mathcal{O}_X$-modules.

http://mathoverflow.net/questions/104422/cohomology-and-j/104430#104430 Andy B 2012-08-10T18:56:14Z 2012-08-10T18:56:14Z

This is great, thanks! For the case I'm interested in, $X$ is not smooth. However, it is a (normal) toric variety. Does this buy me anything?

http://mathoverflow.net/questions/95395/is-a-certain-symmetric-power-reprsentation-of-glm-cyclically-generated/95397#95397 Andy B 2012-04-27T21:46:55Z 2012-04-27T21:46:55Z

$(\mathbb{C}^n)^{\otimes n}$ is cyclically generated by $e_1 \otimes e_2 \otimes \dots \otimes e_n$ and is not multiplicity free.

http://mathoverflow.net/questions/89692/if-graph-is-tree-what-can-be-said-about-its-adjacency-matrix Andy B 2012-02-27T21:30:09Z 2012-02-27T21:30:09Z

The Smith Normal Form of the adjacency matrix must be $(1,\dots,1,0)$.

http://mathoverflow.net/questions/82420/tensor-product-of-universal-highest-weight-modules Andy B 2011-12-02T16:42:24Z 2011-12-02T16:42:24Z

@Chris: You made it sound like they are finite dimensional was all.

http://mathoverflow.net/questions/82420/tensor-product-of-universal-highest-weight-modules Andy B 2011-12-02T16:12:24Z 2011-12-02T16:12:24Z

The modules U, V, and W are not finite dimensional....

http://mathoverflow.net/questions/32099/what-is-lagrange-inversion-good-for Andy B 2011-11-28T22:48:31Z 2011-11-28T22:48:31Z

David, you might look at Mark Haiman's PCMI notes. He uses a q-LI in the context of studying Macdonald polynomials.

http://mathoverflow.net/questions/81109/non-regular-non-coherent-subdivisions-of-a-polygon Andy B 2011-11-17T01:33:24Z 2011-11-17T01:33:24Z

it's a subdivsion that doesn't come from lifting the vertices of the polytope into a higher dimension and then &quot;looking up&quot; to see the facets of the triangulation.