User faisal - MathOverflow

Answer by Faisal for Parallel forms and cohomology of symmetric spaces

2013-04-08T23:38:38Z

I think there's some confusion in the question. For example by "Levi-Civita connection" you must really mean some kind of Laplacian. Anyway, your end result about the cohomology of $G/H$ is essentially correct. To help clarify the situation, let's bring in some relative Lie algebra cohomology.

The first basic fact is that a Lie group $G$ acts naturally on the space $\Omega^\ast(G/H)$ of differential forms on $G/H$, where $H$ is a closed subgroup of $G$. If $G$ and $H$ are connected, then evaluation at the identity gives an isomorphism from the space $\Omega^\ast(G/H)^G$ of $G$-invariant forms onto the space $C^\ast(\mathfrak g, \mathfrak h)$ of relative Lie algebra cohomology cocycles. This isomorphism descends to an isomorphism between the cohomology of the complex of $G$-invariant forms on $G/H$ and relative Lie algebra cohomology $H^\ast(\mathfrak g, \mathfrak h)$.

Now assume that $G$ is reductive and $H$ maximal compact in $G$ with corresponding Cartan decomposition $\mathfrak g = \mathfrak h \oplus \mathfrak p$. Then there are natural isomorphisms $$ C^q(\mathfrak g, \mathfrak h) = \mathrm{Hom}_{\mathfrak h} (\wedge^q \mathfrak p, \mathbb R) = (\wedge^q \mathfrak p)^{\mathfrak h}, $$ where $\mathfrak h$ is acting via the adjoint representation. If you want, this can be rephrased in terms of the action of $H$, and moreover it's possible to introduce a bilinear form so that the above isomorphisms yield $$ C^q(\mathfrak g, \mathfrak h) = (\wedge^q \mathfrak h^\perp)^H. $$ The next key result is that the differential of this complex is zero, whence $$ H^q(\mathfrak g, \mathfrak h) = (\wedge^q \mathfrak h^\perp)^H. $$ In some precise sense this is a manifestation of a theorem of E. Cartan which states that an invariant form on $G/K$ is automatically closed (and harmonic). Indeed, we can combine this last isomorphism with the isomorphisms mentioned in the second paragraph to conclude that $$ \Omega^q(G/H)^G = (\wedge^q \mathfrak h^\perp)^H $$ and both are isomorphic to the cohomology of the complex of $G$-invariant forms on $G/H$.

In general we can't deduce from this that the cohomology of $G/H$ is isomorphic to $(\wedge^\ast \mathfrak h^\perp)^H$, because not every closed form will necessarily have an invariant form in its cohomology class. However, if $G$ is compact, then this is not a problem, because we can use the usual averaging trick.

Anyway, the point of introducing Lie algebra cohomology into the picture is that it makes things very computable. It will also be an important key phrase for any literature search.

Answer by Faisal for Kostant's Theorem on Principal TDS

2013-04-08T17:56:07Z

Is it true that the centralizer $Z_{\frak{g}}(\frak{a})=\{\xi\in\frak{g}:[\xi,\eta]= $0$ \text{ }\forall\eta\in\frak{a}\}$ is trivial (or equivalently, that the trivial one-dimensional representation of $\frak{a}$ is not an irreducible constituent of the $\frak{a}$-module $\frak{g}$)?

The answer is yes. Let's follow Kostant's setup in section 5. Thus, fix a Cartan subalgebra $\mathfrak h$, a set of simple roots $\{\alpha_i\}_{i=1}^r$ , and corresponding root vectors $e_{\alpha_i}$. Then we may assume that $\mathfrak a$ is generated by the $S$-triple $\{h, e,f\}$ as described in the proof of Lemma 5.2 with (say) $e = \sum_{i=1}^r e_{\alpha_i}$. Now, $Z_{\mathfrak g}(\mathfrak a)$ lies inside the centralizer of $h$, which is $\mathfrak h$ by the proof of Theorem 5.2. But if $x \in \mathfrak h$ centralizes $\mathfrak a$ then $$ 0 = [x,e] = \sum_{i=1}^r (x,\alpha_i) e_{\alpha_i} $$ whence $(x,\alpha_i)=0$ for all $i$ and so $x=0$.

Answer by Faisal for Is there an analog of determinant for linear operators in infinite dimensions as that of finite dimensions?

2013-04-04T01:45:53Z

Attempting to define a well-behaved "infinite-dimensional determinant" for all operators will get us into trouble fairly quickly. Consider, for example, a vector space $V$ of countably infinite dimension and let $A$ be multiplication by some nonzero scalar $\lambda\neq1$. Then $A$ is certainly invertible so its putative determinant should be nonzero. But now upon fixing a basis for $V$ we get the following matrix identities $$ A = \begin{pmatrix} \lambda \\ & \lambda \\ & & \lambda \\ & & & \ddots \end{pmatrix} = \begin{pmatrix} \lambda \\ & 1 \\ & & 1\\ & & & \ddots \end{pmatrix} \begin{pmatrix} 1 \\ & \lambda \\ & & \lambda \\ & & & \ddots \end{pmatrix} = \begin{pmatrix} \lambda \\ & I \end{pmatrix} \begin{pmatrix} 1 \\ & A \end{pmatrix}. $$ If we expect our determinant to behave like its finite-dimensional counterpart, then the above would yield $\det A = \lambda \det A$ and consequently that $\det A = 0$, counter to what we expect from the invertibility of $A$.

So in attempting to define $\det$ for operators on infinite-dimensional spaces, you either have to restrict the class of operators under consideration or lower your expectations of how your $\det$ will be analogous to the finite-dimensional one.

Answer by Faisal for On Prufer domains

2013-04-03T22:20:08Z

Yes. In fact, there's an example of such a ring that is also a Bezout domain (so all the finitely generated primes are in fact principal). The construction is as follows. Fix an integer prime $p$, let $\alpha$ be an element of $\mathbb Z_p$ that is transcendental over $\mathbb Q$, and set $A = \{ \alpha + p, \alpha + p^2, \ldots \}$ . The desired ring is then $R = \{ f \in \mathbb Q(X) \colon f(A) \subset \mathbb Z_p \}.$ The nonzero primes of $R$ are all maximal and of the form ${\mathfrak m}_i = \{ f \in R \colon f(\alpha + p^i) \in p\mathbb Z_p\}$ ( $i\geq0$ ). The ideal $\mathfrak m_0$ is not finitely generated but all the others are. This follows from general facts about rings of "integer-valued rational functions" together with the observation that $\alpha$ is a limit point of $A$ while $\alpha + p^i$ (with $i>0$ ) are all isolated points. The relevant details are in Chapter X of Cahen & Chabert, Integer-Valued Polynomials (AMS, 1997); see exercises 19 and 20 in particular.

Incidentally, the localization of $R$ at each $\mathfrak m_i$ is a dvr, but $R$ is not noetherian because ${\mathfrak m}_0$ is not finitely generated. Thus $R$ is an example of an "almost Dedekind domain" that isn't Dedekind. A question about such rings came up not too long ago.

Answer by Faisal for Moduli Spaces of Higher Dimensional Complex Tori

2013-04-03T02:19:39Z

The fact that all $1$-dimensional tori are projective means care is sometimes needed in making analogies with higher dimensional tori. This is one of those times. The natural 'moduli space' of all $d$-dimensional complex tori constructed by Will is not very nice when $d>1$. For example, the action of $GL_{2d}(\mathbb Z)$ on $GL_d(\mathbb C)\backslash GL_{2d}(\mathbb R)$ isn't properly discontinuous, so the resulting quotient space isn't Hausdorff (an observation due to Siegel).

A slightly better state of affairs is available if we begin by looking at the Siegel upper half-plane $$ \mathcal H_d = \{ \tau \in M_d(\mathbb C) \colon \tau^t = \tau, \mathrm{Im}(\tau) > 0 \}. $$ The group $Sp(2d,\mathbb R)$ acts transitively on $\mathcal H_d$ via $$ \begin{pmatrix}A & B\\C & D\end{pmatrix} \tau = (A\tau+B)(C\tau+D)^{-1} $$ (this is well-defined) and the isotropy subgroup of $iI_d$ is (essentially) $U(g)$. Thus we can view $\mathcal H_d$ as $Sp(2d,\mathbb R)/U(d)$. Now a point $\tau \in \mathcal H_d$ gives us a complex torus $\mathbb C^d/(\mathbb Z^d + \mathbb Z^d \tau)$ -- but this torus is not an arbitrary one: it comes with a "principal polarization". Moreover the natural action of $Sp(2d,\mathbb Z)$ on $\mathcal H_d$ (which, by the way, is properly discontinuous) preserves the isomorphism class of the corresponding torus as well as its principal polarization. Thus the points of the space $\mathcal H_d/Sp(2d,\mathbb Z)$ are in bijection with isomorphism classes of principally polarized $d$-dimensional complex tori (these are abelian varieties). Note that if $d=1$ we're reduced to the familiar modular curve $\mathcal H/SL(2,\mathbb Z)$ which parametrizes all $1$-dimensional tori.

For (much) more on this, see II.4 of Mumford's Tata Lectures on Theta I or van der Geer's chapter in The 1-2-3 of Modular Forms.

Answer by Faisal for Are there any nontrivial ring homomorphisms $M_{n+1}(R)\rightarrow M_n(R)$?

2013-03-31T21:30:04Z

We can also rule out the case of commutative $R$ by appealing to the Artin–Procesi theorem: an Azumaya algebra of constant rank $(n+1)^2$ (e.g. $M_{n+1}(R)$) satisfies all the $\mathbb Z$-multilinear identities of $M_{n+1}(\mathbb Z)$ but no nonzero homomorphic image of it satisfies all the $\mathbb Z$-multilinear identities of $M_n(\mathbb Z)$.

It's perhaps worth noting that if R is a field, then there's a fairly straightforward way of proving that there is no injective ring homomorphism M_{n+1}(R) \to M_n(R). In fact, suppose we have a nonzero ring homomorphism M_{n'}(R) \to M_n(R). Then this allows us to view R^n as a left M_{n'}(R)-module. Now if R is a field, then M_{n'}(R) is simple, and so R^n decomposes into a finite direct sum of irreducible M_{n'}(R)-modules. It's a standard fact (and one that is easy to prove) that each such module is isomorphic to R^{n'}. We thus obtain an isomorphism R^n = R^{n'} \oplus \cdots \oplus R^{n'} of M_{n'}(R)-modules, and hence of R-vector spaces by restricting the action to the subring of scalar matrices. But then linear algebra allows us to conclude that n'|n. Nevermind. :)

Update: It's possible to have a nontrivial ring map $M_{n+1}(R) \to M_n(R)$ with $R$ finitely generated (and necessarily noncommutative). The idea, inspired by my previous mishap and wccanard's comment, is to find a finitely generated ring $R$ for which there is an isomorphism $R^{n+1} \cong R^n$ of left $R$-modules. In this case one obtains ring isomorphisms $$ M_{n+1}(R) \cong \mathrm{End}_R(R^{n+1}) \cong \mathrm{End}_R(R^n) \cong M_n(R). $$ The ring theorists provide us with examples of such rings. In fact, for any positive integers $n < m$, Leavitt gives a finitely generated ring $L_{n,m}$ for which there is a left $L_{n,m}$-module isomorphism $L_{n,m}^n \cong L_{n,m}^m$ and, consequently, a ring isomorphism $M_n(L_{n,m}) \cong M_m(L_{n,m})$.

Answer by Faisal for Character fields and Clifford's theorem

2013-03-17T17:32:01Z

I don't think there's any clear cut relationship. In particular, the inequality $e \ge [\mathbb Q(\eta) : K]$ needn't hold. For example, take $G=S_5$, $N=A_5$ and let $\chi$ be the unique irreducible $S_5$-character of degree $6$. Then we have $\mathrm{res}^{G}_{N} \chi = \eta + \bar{\eta}$, where $\eta$ and $\bar{\eta}$ are the degree $3$ irreducible characters of $A_5$. So here $e=1$ and $\mathbb Q(\eta) = \mathbb Q(\sqrt 5)$. On the other hand, $\chi$ is integer valued so $K=\mathbb Q$. Thus $e < [\mathbb Q(\eta) : K]$ in this case.

Answer by Faisal for Exercise in Milne's CFT notes

2013-02-24T01:50:09Z

How did you determine that the index $(\mathcal O_L : M)$ is $8$? It seems to me that it's actually $160$, which is divisible by $5$.

Incidentally, a quick way to see that Milne is correct is to note that the discriminant of $K$ is $-4\cdot 6$ and $6$ is an idoneal number: this means that the Hilbert class field of $K$ coincides with its genus field, which is easily computed to be $K(\sqrt{-3})$. See section 6 of Cox's "Primes of the form $x^2 + ny^2$".

Edit: As requested, here are some more details, as well as a low-tech way of seeing why $L=K(\sqrt{-3})$ is the Hilbert class field of $K$.

First off, an integral basis for $L$ is $\{1, (1+\sqrt{-3})/2, \sqrt2, (\sqrt 2 + \sqrt{-6})/2\}$ (taken from an exercise in Marcus's "Number Fields"), and then a simple computation gives $\text{disc } \mathcal O_L = 12^2$. On the other hand, we find that $\text{disc } M = 1920^2$ and so the equation $\text{disc } M = (\mathcal O_L : M)^2 \text{disc } \mathcal O_L$ gives $(\mathcal O_L : M) = 160$.

Now, as to why $L$ is the Hilbert class field of $K$---well, the HCF has to be a quadratic extension of $K$ (because the class number of $K$ is easily computed to be $2$), so it suffices to show that the finite primes of $K$ are unramified in $L$. For this we can use the relative discriminant of $L/K$: this is an ideal that contains, in particular, $\text{disc } \{1,\sqrt 2\} = 8$ and $\text{disc } \{1, (1+\sqrt{-3})/2\} = -3$ , hence contains $1$. That is, the relative discriminant of $L/K$ is the unit ideal and so the extension is unramified.

Answer by Faisal for Reconstructing a Lie group Banach representation from the Lie algebra rep. on analytic vectors

2013-02-14T19:00:20Z

In general you can't expect to get an action of $G$ on $V^\omega$. Instead, what one has is that if $U$ is a $U(\mathfrak g_{\mathbb C})$-invariant subspace of $V^\omega$ then its closure $\overline{U}$ will be $G$-invariant. This fact makes $V^\omega$ particularly useful---for example, the analogous statement is false for the space $V^\infty$ of smooth vectors. In any case, by applying this to $U=V^\omega$ and using the density theorem $\overline{V^\omega} = V$ mentioned by Jim Humphreys, we recover the $G$-action on all of $V$.

Answer by Faisal for Is a reductive adelic group a Type I group?

2012-12-13T21:08:50Z

I believe the answer is yes. Let's begin by recalling that if one wants to show that a locally compact group $G$ is of type I, it suffices to show that $G$ contains a "large" compact subgroup $K$, in the sense that for every $\pi \in \hat{G}$ and $\sigma \in \hat{K}$, the multiplicity of $\sigma$ in $\pi|_K$ is finite. This is how Harish-Chandra showed that a real reductive group is of type I (take $K$ to be a maximal compact), and also how Bernstein showed that a $p$-adic reductive group is of type I (take $K$ to be a compact open subgroup).

Now let $G$ be a connected reductive group over $\mathbb Q$. Then, away from a finite set $S$ of places (containing $\infty$), $G$ is unramified and has a model over $\mathbb Z_p$. Let's abuse notation and denote this model by $G$. It suffices to show that $G(\mathbb A^S) = \prod'_{p \not\in S} G(\mathbb Q_p)$ is of type I. The desired large $K$ turns out to be $K = \prod_{p \not\in S} G(\mathbb Z_p)$. This assertion essentially appears (without proof) as Theorem 4 in Flath's article in the Corvallis proceedings. The details are spelled out in the appendix to Clozel's article in the IAS/Park City 2002 lecture notes on automorphic forms (MR2331351; a Google Books preview is available here).

Answer by Faisal for Is a domain all of whose localizations are noetherian itself noetherian ?

2012-11-28T00:46:06Z

I had the exact same question not too long ago. Apparently if you drop the noetherian precondition in Neukirch's definition of "Dedekind domain" then you get what some people call an "almost Dedekind domain". There are indeed examples of almost Dedekind domains that aren't Dedekind (i.e. aren't noetherian). The first of these was given by Nakano (J. Sci. Hiroshima Univ. Ser. A. 16, 425–439 (1953)): take the integral closure of $\mathbb Z$ in the field obtained by adjoining to $\mathbb Q$ the $p$th roots of unity for all primes $p$.

Answer by Faisal for Computing determinants of characters

2012-11-27T00:56:24Z

If you know $\chi$ then you can write down $\det \chi$ using Newton's identities. This is simply the observation that one can express the determinant of a matrix $\rho(g)$ in terms of traces of powers $\rho(g)^k=\rho(g^k)$ of that matrix.

Answer by Faisal for Product of conjugacy classes - is there an analog of Tanaka-Krein reconstruction ?

2012-10-27T22:07:42Z

The answer to your first question is negative. For a concrete example, you can show that the conjugacy class rings of the nonisomorphic groups $Q_8$ and $D_8$ are isomorphic, via an isomorphism that pairs off the bases as follows: $[1] \leftrightarrow [1]$, $[-1] \leftrightarrow [r^2]$, $[i] \leftrightarrow [r]$, $[j] \leftrightarrow [s]$ and $[k] \leftrightarrow [rs]$.

As to your question about the relationship between the conjugacy class ring and the character ring, there are lots of partial results that can be stated. Nonetheless, the answer to the question of when these two rings are isomorphic is completely known. This turns out to be the case if and only if the group is $p$-nilpotent with abelian Sylow $p$-subgroup. More generally, for arbitrary finite groups $G$ and $G'$, the following two conditions are equivalent.

The character ring of $G$ is isomorphic to the conjugacy class ring of $G'$.
$G$ and $G'$ are $p$-nilpotent groups with abelian Sylow $p$-subgroups. Moreover, if $g_1, \dots, g_l$ and $g_1',\ldots, g_{l'}'$ are complete sets of representatives for the conjugacy classes of $p'$-elements of $G$ and $G'$, resp., and if $D_i$ and $D_i'$ are Sylow $p$-subgroups of $C_G(g_i)$ and $C_{G'}(g_i')$, resp., then $l=l'$ and $D_i \cong D_i'$.

This is due to Saksonov, The ring of classes and the ring of characters of a finite group. Mat. Zametki 26 (1979), no. 1, 3–14, 156.

Answer by Faisal for How to calculate the equivariant cohomology ring of $P^2$?

2012-10-25T18:16:00Z

It sounds like you're talking about GKM (=Goresky–Kottwitz–MacPherson) theory, in which case it's better to think of tori as complex tori, i.e. as products of copies of $\mathbb C^\times$ and not of $S^1$. The triangle to which you're referring is the so-called moment graph of $\mathbb{CP}^2 = SL_3(\mathbb C)/P$, where $$ P = \begin{pmatrix} \ast & \ast & \ast \\ 0 & \ast & \ast \\ 0 & \ast & \ast \end{pmatrix} $$ and $T=\mathbb C^\times \times \mathbb C^\times$ is the diagonal subgroup of $SL_3$ acting by left multiplication. The vertices of the moment graph are the $T$-fixed points, of which there are three in this case. Two vertices are connected by an edge if and only if there's a one-dimensional $T$-orbit whose closure contains the corresponding fixed points. The closure of such an orbit is a copy of $\mathbb{CP}^1$, so that might explain why your professor labeled the edges as such. But anyway, the moment graph already comes with a useful labeling and a direction, though let me not say more about this here.

GKM theory provides a combinatorial description of $H_T^\ast(M)$ in terms of the moment graph of $M$, and it appears that this is what your professor was using. (Here $M$ refers to a projective variety on which a complex torus $T$ is acting in some "nice" fashion. If $M=G/P$ is a generalized flag variety then the action of a maximal torus $T\subset P$ of $G$ is "nice" enough for GKM theory.) One is also provided with an isomorphism $$ H^\ast(M) = \frac{H_T^\ast(M)}{\mathfrak{m} H_T^\ast(M)}, $$ where $\mathfrak m$ is the augmentation ideal $(x_1,\ldots,x_n)$ in $H_T^\ast({\text pt}) \cong S(\mathfrak{t}^\ast) \cong \mathbb C[x_1,\ldots,x_n]$.

For more on this, I recommend Julianna Tymoczko nice survey article. Be sure to check out example 4.1, where she computes $H_T^\ast(\mathbb{CP}^2)$ for our $T$ above.

Answer by Faisal for Heegner Points and Binary Quadratic Forms

2012-10-22T19:38:26Z

I think the following facts, which you can find in Cox's book Primes of the Form $x^2+ny^2$, will alleviate your confusion. First off, if ${\mathfrak a}=[\alpha,\beta]$ is a proper ideal of ${\mathcal O}$ then one can show that $$ f(x,y) := \frac{N(\alpha x-\beta y)}{N{\mathfrak a}} $$ is a primitive binary quadratic form of discriminant $D = {\rm disc}(\mathcal O)$ . Moreover, the map that associates such an ${\mathfrak a}$ to such an $f(x,y)$ induces an isomorphism from the class group ${\rm Pic } ({\mathcal O})$ onto the form class group $C(D)$ . The inverse of this map is given by $$ f(x,y) := ax^2+bxy+cy^2 \mapsto [a,(-b+\sqrt{D})/2] = [a,a\tau], $$ where $\tau$ is the unique point in the upper-half plane such that $f(\tau,1)=0$ . It's not hard to show that we'll have ${\mathcal O} = [1,a\tau]$ for all such $\tau$ (see Addendum below). In particular, we see that ${\mathcal O}/[a,a\tau] \cong {\mathbb Z}/a{\mathbb Z}$ is cyclic.

The last piece of the puzzle is this: a positive integer $N$ is represented by a form $f(x,y)$ in $C(D)$ if and only if $N$ is the norm of some ideal in the corresponding ideal class in ${\rm Pic}({\mathcal O})$ (loc. cit., Theorem 7.7(iii)). On the other hand, $N$ is properly represented by such an $f(x,y)$ if and only if $f(x,y)$ is properly equivalent to $Nx^2+bxy+cy^2$ for some $b,c \in {\mathbb Z}$ . Now the results mentioned in the preceding paragraph will take you home.

Addendum: Given a proper ideal $\mathfrak a$ of $\mathcal O$ , we can recover $\mathcal O$ as the set ${\mathfrak a}^\vee = \{x \in K \mid x\mathfrak a \subset \mathfrak a \}$ . This last set is easy to compute in the following special case. Let $K=\mathbb Q(\tau)$ be quadratic and suppose that $ax^2+bx+c$ is the minimal polynomial of $\tau$ , where $a,b,c$ are coprime integers. Then $[1,\tau]^\vee = [1,a\tau]$ (loc. cit., Lemma 7.5).

By applying this to Gross's $\mathfrak a = [\omega_1, \omega_2] = \omega_2 [\tau, 1]$, which is a proper ideal in some order $\mathcal O$ , we find that $\mathcal O = [A\tau, 1]$ . Consequently,$$D = {\rm disc}({\mathcal O}) = \det \begin{pmatrix}1 & A\tau \\ 1 & A\bar{\tau} \end{pmatrix}^2 = B^2 - 4AC.$$ The assertion about $A$ can be gotten in a similar manner.

Answer by Faisal for Does there exist a non effective divisor with positive degree?

2012-10-14T04:54:18Z

Here's one way of seeing why Piotr's claim is true. Write $g$ for the genus of $X$ and begin with the following observation. If $p_1, \ldots, p_g$ are points on $X$ , then the divisor $E=p_1+\cdots+p_g$ has $l(E) = g + 1 - \rho$ , where $\rho$ is the rank of the associated Brill–Noether matrix. This matrix is $g\times g$ so for points $p_1, \ldots, p_g$ in general poisiton, we'll have $\rho=g$ and consequently $l(E)=1$ . (For a more precise statement, see §7c in Gunning, Lectures on Riemann Surfaces, PUP 1966.) Now take one such $E$ and consider the divisor $D=E-q$ . We have $\deg D = g-1 > 0$ and, for generic $q$ , $l(D)=l(E)-1=0$ .

If $g=0$ or $1$ then it follows easily from Riemann–Roch that $\deg D > 0 \implies l(D)>0$ .

Answer by Faisal for References request on the algebraic geometry of projective homogeneous spaces

2012-10-12T00:45:48Z

In addition to the references already mentioned, let me recommend Dennis Snow's excellent notes on homogeneous vector bundles. They cover everything you ask for (and much more):

For a description of ${\rm Pic}(G/P)$ in terms of weights, see Theorem 6.4.
Ample line bundles are characterized in Theorem 6.5.2.
The canonical class is described on page 37, right above Definition 10.2.

Answer by Faisal for The anticanonical bundle on a flag variety is ample

2012-06-14T00:09:28Z

Let me try to add to Jim Humphreys's comprehensive answer by pointing out that the (very) ampleness of line bundles coming from regular dominant weights was already observed by Borel and Weil, and essentially appears as Thm.4 in Serre's Bourbaki report

Représentations linéaires et espaces homogènes kählériens des groupes de Lie compacts (d'après Armand Borel et André Weil). Séminaire Bourbaki, Vol. 2, Exp. No. 100, 447–454, Soc. Math. France, Paris, 1995.

(See the example following Thm.4 on p.453.)

Although Borel and Weil work over $\mathbb C$ their argument is probably worth summarizing here. They begin by observing that the line bundle $L_\lambda \to G/B$ coming from a weight $\lambda$ is spanned by global sections iff $\lambda$ is dominant. In this case the image of the map $G/B \to \mathbb P (H^0(G/B,L_\lambda)^\ast)$ defined by sections is of the form $G/P$ where $P\supset B$ is the parabolic subgroup defined by the simple roots perpendicular to $\lambda$ . Thus if $L_\lambda$ is not very ample, so that the aforementioned fibration $G/B \to G/P$ is not an embedding, one concludes that $P$ must properly contain $B$ and consequently that $\lambda$ must be perpendicular to some simple root, hence is not regular.

Answer by Faisal for Which formulae of Euler is Fröhlich referring to?

2012-05-07T14:30:18Z

I believe the reference is to this formula of Euler (see here): If $P(x)/Q(x)$ is a rational function and $ax+b$ is a simple factor of $Q(x)$, then the coefficient of $1/(ax+b)$ in the partial fraction decomposition of $P/Q$ is given by $$ \lim_{x\to \frac{-b}{a}} \frac{a P(x)}{Q'(x)}. $$

To see how this applies here, proceed as Serre does in Local Fields. That is, write (in a suitable extension of $L$) $$ \frac{1}{g(X)} = \sum_{k=1}^n \frac{a_k}{X-x_k} \qquad (*) $$ where the $x_k$ are the conjugates of $x$. Then a formal application of Euler's formula gives $$ a_k = \lim_{X \to x_k} \frac{1}{g'(X)} = \frac{1}{g'(x_k)}. $$ Now expand both sides of (*) as power series in $1/X$ and compare coefficients.

Answer by Faisal for connected compact semisimple lie group finite fundamental group

2012-05-01T14:51:28Z

There is a quick proof via Lie algebra cohomology: Let $G$ denote your compact, connected, semisimple Lie group, and let $\mathfrak g$ denote its Lie algebra. Then $$ H^1(G;\mathbb R) = H^1(\mathfrak g;\mathbb R) = \text{Hom}_{\mathbb R} (\mathfrak g/[\mathfrak g, \mathfrak g], \mathbb R) = 0, $$ whence $H_1(G;\mathbb Z)$ is finite. But $\pi_1(G)$ is abelian hence is isomorphic to $H_1(G;\mathbb Z)$. QED.

Answer by Faisal for Orthogonal subgroups of dual group

2012-01-18T02:29:05Z

Lemma 2.1.3 in Rudin's Fourier Analysis on Groups does this for locally compact abelian $G$ and closed $H \leq G$. This might not the best reference (may be a bit too general for a CS readership?), but it was the first one to pop into my head.

Answer by Faisal for real orbits of highest weight vectors

2011-11-09T22:19:04Z

There seems to be some confusion in the question, so let me try to recap the basic setup. Thus let $G$ be a complex simple Lie group, $V^\lambda$ the irrep of $G$ of highest weight $\lambda$, and $G/P$ the $G$-orbit in $\mathbb P V^\lambda$ through the line spanned by a highest weight vector in $V^\lambda$. In the question it's claimed that if $G_0$ is a real form of $G$ then the action of $G_0$ on $G/P$ is transitive. This is not true. For example, if $G=\operatorname{SL}_2 \mathbb C$ then the action of $G_0 = \operatorname{SL}_2 \mathbb R$ on the sphere $G/P = \mathbb P^1$ has three orbits (two hemispheres and the circle in between them). In general, there will be finitely many $G_0$-orbits in $G/P$. A wealth of information can be found in

Joseph Wolf, The action of a real semisimple group on a complex flag manifold. I: Orbit structure and holomorphic arc components, Bull. Amer. Math. Soc. 75 (1969), 1121-1237.

If $G_0$ is a compact real form then there is only one orbit, i.e. $G_0$ acts transitively $G/P$. This gives the familiar "compact picture" $G_0/G_0 \cap P$ of the flag variety. On the other hand, if $G_0$ is noncompact then it's relatively uncommon for its action on $G/P$ to be transitive, but it is possible. Joseph Wolf worked out the list of $G_0$ for which this is the case in

Joseph Wolf, Real groups transitive on complex flag manifolds, Proc. Amer. Math. Soc. 129 (2001), 2483-2487.

Answer by Faisal for A technical problem on the contragredient representation in the context of locally compact totally disconnected groups

2011-11-09T18:51:56Z

This follows from two facts:

The complement $E_1^\perp$ of $E_1$ in $\tilde{E}$ is isomorphic to the contragredient of $E/E_1$.
If $V$ is admissible and nonzero then $\tilde{V}$ is nonzero (and admissible). For if $\tilde{V}=0$ then $V = \tilde{\tilde{V}} = 0$.

Answer by Faisal for Many p,q-Sylow subgroups

2011-11-06T19:39:56Z

The answer is no: see Corollary 1.3 in

Robert M. Guralnick; Gunter Malle; Gabriel Navarro, Self-normalizing Sylow subgroups, Proc. Amer. Math. Soc. 132 (2004), 973-979.

Answer by Faisal for What is significant about the half-sum of positive roots?

2011-11-06T00:11:03Z

This is actually a fairly deep question. Your suspicion that there may be multiple answers is correct, but there might be some surprising connections between seemingly unrelated answers. Let me give one possible thread of explanation. The underlying principle is that the appearance of $\rho$ and the "dot" action $w\cdot\lambda=w(\lambda+\rho)-\rho$ in representation theory is closely related to the geometry of the flag variety.

One of the first places one meets $\rho$ (and the dot action) is in the Weyl character formula. A theorem of Kostant shows that the formula can be written as the ratio of two Lie algebra cohomology Euler characteristics. From this perspective, the appearance $w \cdot \lambda$ and $w\cdot0$ in the WCF is ultimately explained by the fact that these are the weights appearing in the weight space decomposition of the relevant Lie algebra cohomology modules, namely $H^*(\mathfrak n, V^\lambda)$ and $H^\ast(\mathfrak n, V^0)$ , where $\mathfrak n = \bigoplus_{\alpha>0} \mathfrak g_\alpha$ and $V^\mu$ denotes the irrep of highest weight $\mu$ .

We can rephrase this in geometric terms by invoking the "geometric analogue" of Kostant's theorem, i.e. the Borel–Weil–Bott theorem. Kostant's description of the Lie algebra cohomology of $\mathfrak n = \mathfrak g /\mathfrak b^-$ with coefficients in an irrep translates into a representation-theoretic description of the sheaf cohomology of certain line bundles $L_\lambda$ (constructed using integral weights $\lambda$ ) over the flag variety $G/B^-$ of $\mathfrak g$ . Consequently, the dot action shows up in this description, and this time it's accompanied by a shift in degree. This in turn can be explained by Serre duality; the key fact is that canonical bundle of $G/B^-$ turns out to be $L_{-2\rho}$ .

So, in some sense, the appearance of $\rho$ and the dot action in the WCF can be thought of as a manifestation of Serre duality.

[N.B. This is a condensed version of my lengthy original answer. The old version can be found in the edit history.]

Answer by Faisal for Eigenvalues of Krylov matrices

2011-11-06T04:25:30Z

I don't see any reason for there to be a nice characterization. For instance if $A$ is diagonal then $K_n$ is a Vandermonde matrix, so its spectrum is fairly complicated...

Answer by Faisal for Does every finite nontrivial group have two distinct irreducible representations over the complex numbers of equal degree?

2011-10-30T02:33:41Z

It seems that the answer is yes. A MathSciNet search brought up the paper

Y. Berkovich, D. Chillag, and M. Herzog, Finite groups in which the degrees of the nonlinear irreducible characters are distinct, Proc. Amer. Math. Soc. 115 (1992), 955–959.

In it you can find a characterization of groups whose nonlinear irreducible characters have distinct degrees. In particular, such a group can't be perfect (see Lemma 1), and so will always have multiple linear characters as was noted in the OP. The proof, however, relies on the classification of finite simple groups, so is not "easy".

Addendum: I took a closer look at the related literature and happened across the following interesting result, which I figured was worth sharing. (It can also be used to give an affirmative answer to the original question.)

Theorem. Let $G$ be a nontrivial finite group. If the character table of $G$ has a column or row containing distinct rational entries, then $G$ must be isomorphic to either $S_2$ or $S_3$ .

The reference is

M. Bianchi, D. Chillag, A. Gillio, Finite groups with many values in a column or a row of the character table, Publ. Math. Debrecen 69 (2006), no. 3, 281–290.

The result from the classification of finite simple groups used in the Berkovich–Chillag–Herzog paper is also used here (in very much the same spirit).

Answer by Faisal for T-bundles and the Borel-Weil-Bott theorem

2011-10-28T08:13:05Z

I'm very skeptical about the possibility of getting the full Borel–Weil–Bott theorem just by studying $G/U \to G/B$ . Probably the closest thing I can think of is Bott's original proof of his theorem, which involves studying certain $\mathbb P^1$ -bundles $G/B \to G/P$ . On the other hand, you can prove the Borel–Weil theorem by studying the function space $\mathcal{O}(G/U)$ , but even here you need to know a little more than just that this space contains every irrep of $G$ exactly once. More specifically, you want to know how each irrep shows up. Let me sketch the argument. To be safe, I assume we're working over $\mathbb C$ , but what follows probably works over any algebraically closed field of characteristic zero.

To start off, note that $G$ acts on $\mathcal{O}(G)$ by left and right translation. Viewing $\mathcal{O}(G)$ under the latter action, we can think of $$ \mathcal{O}(G/U) = \{ f \in \mathcal{O}(G) \colon f(gu) = f(g) \text{ for all } g \in G, u \in U \} $$ as the space $\mathcal{O}(G)^U$ of $U$ -invariants. Now recall that there's a $G\times G$ -equivariant decomposition $$ \mathcal{O}(G) = \bigoplus V \otimes V^\ast \qquad \text{[an algebraic Peter–Weyl theorem]}$$ where the sum runs over the irreps of $G$ , and $G$ acts on $V$ by left translation and on $V^\ast$ by right translation. Therefore we find that $$ \mathcal{O}(G/U) = \mathcal{O}(G)^U = \bigoplus V \otimes (V^\ast)^U. $$ Let's assume that $U$ is built up using negative roots, so that $(V^\ast)^U$ is the lowest weight space of $V^\ast$ , and in particular is one-dimensional. This shows that every irrep of $G$ appears in $\mathcal{O}(G/U)$ exactly once. But that's not all: using the right $G$ -action, we can "capture" the irrep of highest weight $\lambda$ . Indeed, as a $T$ -module, $(V^\ast)^U = \mathbb C_\mu$ , where $\mu$ is the lowest weight of $V^\ast$ , or said differently, $-\mu$ is the highest weight of $V$ . So, using the fact that $\text{Hom}_T(\mathbb C_\lambda, \mathbb C_\mu) = \delta_{\lambda\mu} \mathbb C_\lambda$ , we see that the irrep of $G$ of highest weight $\lambda$ can be gotten as $$ \text{Hom}_T(\mathbb C_{-\lambda}, \mathcal{O}(G/U)) = \bigoplus V \otimes \text{Hom}_T(\mathbb C_{-\lambda}, (V^\ast)^U). $$

We can re-write the left side of the above as $$\begin{align} (\mathbb C_\lambda \otimes \mathcal{O}(G/U))^T &= \{ f \in \mathcal{O}(G) \colon f(gtu) = \lambda(t)^{-1} f(g) \text{ for all } g \in G, t \in T, u \in U \} \\ &= \{ f \in \mathcal{O}(G) \colon f(gb) = \lambda(b)^{-1} f(g) \text{ for all } g \in G, b \in B \}, \end{align}$$ which of course we can think of as the space of global sections of the line bundle $L_\lambda = G \times_\lambda \mathbb C$ over $G/B$ . This proves the first part of the Borel–Weil theorem, namely that if $\lambda$ is dominant then $H^0(G/B,L_\lambda)$ is the irrep of highest weight $\lambda$ . The other part, that $H^0(G/B,L_\lambda)=0$ if $\lambda$ is not dominant also follows easily. Indeed, all of the above works just as well for such $\lambda$ , except in this case we have $\text{Hom}_T(\mathbb C_{-\lambda}, (V^\ast)^U)=0$ for all irreps $V$ .

Answer by Faisal for Reversed disc algebra?

2011-10-23T15:07:03Z

No. The spectrum of $A(U)$ is $\overline{U}$ , while the spectrum of the disc algebra is $\overline{\mathbb D_1}$ , and these two spaces aren't homeomorphic.

Answer by Faisal for Complemented subspaces of Banach spaces

2011-10-22T02:09:54Z

I believe the answer to your first question is no. The counterexample I have in mind is related to the peculiar fact (first proved by Enflo, Lindenstrauss and Pisier) that being isomorphic to a Hilbert space isn't a "three-space property". More specifically, there is an example of a Banach space $E$ with an uncomplemented subspace $F$ such that both $F$ and $E/F$ are isomorphic to $\ell^2$ . You can also find examples where $F$ and $E/F$ are isomorphic to any $\ell^p$ . This is due to Kalton and Peck.

Comment by Faisal

2013-03-31T22:31:53Z

Ah, very good point. It seems like the only way to salvage the argument is to assume that the ring map is actually a map of $R$-algebras, but in this case there's a much easier argument... Oh well.

Comment by Faisal

2013-03-31T21:47:48Z

@Dag Oskar Madsen: Your comment is on point! If $R$ is commutative then the kernel of any ring hom $f \colon M_{n+1}(R) \to M_n(R)$ is of the form $M_{n+1}(I)$ for some ideal $I$ of $R$ and so we get an injective map $M_{n+1}(R/I) \cong M_{n+1}(R)/\ker f \to M_n(R)$, which forces $f$ to be 0 by A-L.

Comment by Faisal

2012-11-28T02:02:44Z

@Filippo: That can't be right. The localization of that ring at any any nonzero prime $\mathfrak p$ is a dvr, and this forces the ring to be one-dimensional.

Comment by Faisal

2012-10-22T23:37:27Z

I added some clarification. I hope it helps.

Comment by Faisal

2012-06-13T23:03:49Z

Piotr: Yes, 1) essentially proves the fundamental theorem of algebra.

Comment by Faisal

2012-05-08T16:08:04Z

No problem! $ $

Comment by Faisal

2012-05-02T18:00:29Z

Do you know for a fact that this was meant as a joke (e.g. did you hear this from Lang)? I don't think it's a joke at all; rather it's just a standard application of Minkowski's bound to the computation of the class number of a number field, and a standard usage of the phrase "ring of integers" to refer to $\mathfrak{o}_k$ and not $\mathbb Z$.

Comment by Faisal

2012-05-02T17:15:20Z

I think you might've misinterpreted Lang's usage of the phrase "ring of integers": his argument shows that the ring of integers of $k$ (i.e. $\mathbb Z[\alpha]$) is a PID.

Comment by Faisal

2012-05-02T01:04:39Z

The first 'equality' follows from the observation that the complex of left invariant forms computes $H^\ast_{\rm dR}(G)$ (not difficult to prove) together with the fact that said complex can be identified (via evaluation at the identity) with the complex $\wedge^q\mathfrak g^\ast$. You can find all the details in, e.g., Chevalley--Eilenberg.

Comment by Faisal

2012-05-02T01:03:49Z

Depends on how you set things up. E.g. let's view $H^\ast(\mathfrak g;\mathbb R)$ as being computed by the complex $(\wedge^q\mathfrak g^\ast,d)$ (I'll omit the formula for $d$...). Then the second equality is trivial: $f\in\mathfrak g^\ast$ is in $H^1$ iff $df=0$ (there are no 1-coboundaries). The formula for $d$ here is $df(X,Y)=f([X,Y])$, whence $f\in H^1 \iff f\in (\mathfrak g/[\mathfrak g,\mathfrak g])^\ast$.

Comment by Faisal

2012-05-01T15:18:10Z

By the way, this argument is most likely due to Chevalley--Eilenberg. I would give a more precise reference but my erratic internet connection is making this difficult.

Comment by Faisal

2012-04-07T01:03:53Z

The key observation is that in this case $b_1(X)$ is even, say equal to $2g$. So if $\pi_1(X)$ were free, it would be free of rank $2g$, and would therefore contain a subgroup $H$ of index 2, which is itself free of rank $|G:H|(\text{rank }\pi_1(X)-1)+1=4g-1$. But now consider the 2:1 covering $\tilde{X}\to X$ corresponding to $H$: it has $b_1(\tilde{X})=4g-1$, which is odd. This is a contradiction because $\tilde{X}$ must be compact Kahler as well.

Comment by Faisal

2012-04-06T23:35:50Z

It should be pointed out that this answer, with little modification, could also be used to show that $\pi_1$ of any compact Kahler manifold is never free (unless it's trivial); cf. Mohan Ramachandran's comment.

Comment by Faisal

2012-04-06T23:31:07Z

I miss the days when I could idly peruse MathOverflow and learn a lot of interesting math from great answers to "low level" questions such as this one. Even though I was able to find an answer to this specific question, I still managed to learn quite a bit from all the other nice answers---about things I doubt I would have come across otherwise. It was this sort of experience that attracted me to MO in the first place, and I lament the fact that it is now rarer to come by. Anyway, thank you, Georges, for asking this question.

Comment by Faisal

2012-01-12T18:21:28Z

I don't think "Schur's lemma" is completely obvious in this setting either, especially when $G = GL_n \mathbb R$ is disconnected.