User wilberd van der kallen - MathOverflow

Answer by Wilberd van der Kallen for How to think about parabolic induction.

2012-12-10T13:23:21Z

Let me contribute some confusion. In the situation I am familiar with (algebraic groups), one may first restrict to a Borel subgroup $B$ with the property that $B\cap L$ is a Borel subgroup of $L$. Recall that if $P$ contains both $L$ and $B$, then restricting from $P$ to $B$ and next inducing back up to $P$ does nothing to $P$-modules. So instead of inducing up from $P$ one might as well first restrict to $B$ and then induce up from $B$ to $G$. And all Borel subgroups are conjugate.

Answer by Wilberd van der Kallen for Are homogeneous components of f.g graded modules f.g ?

2012-09-18T18:29:53Z

The answer is yes. Instead of showing that $M_n$ is finitely generated we may show it has the property that any ascending sequence of $A_0$-submodules stabilizes. If $N$ is an $A_0$-submodule of $M_n$, consider $M_n\cap AN$. It is a sum of the $M_n\cap NA_i$. One sees it is $N$ itself. So $N$ can be recovered from $AN$. Now if $N_1\subset N_2\dots$ is an ascending sequence of $A_0$-submodules of $M_n$, the ascending sequence $AN_1\subset AN_2\dots$ stabilizes, hence so do the $N_i=M_n\cap AN_i$.

Answer by Wilberd van der Kallen for Comparing Spectral Sequences

2012-08-15T20:04:19Z

The cokernel is entirely due to $\bar{E}^\infty_{*,0}$ but the kernel is more mysteriuous.

First observe that if $g:C_\cdot\to D_\cdot$ is a chain map and $i$ is such that $g_i$ is surjective and $g_{i-1}$ is injective, then $H_i(g)$ is surjective and $H_{i-1}(g)$ is injective. (Exercise.)

Using this, one sees by induction on $r$ that $f^r_{p,q}$ is surjective for $q\geq1$ and injective (hence bijective) for $q\geq r-1\geq1$.

Now take $r=\infty$. One sees that $h$ hits all layers except the top one in the filtration of $\bar{H}_*$.

Let me include Grilo's formulas as I now believe they should read: We have exact sequences

$$0\to{\bar{E}}^{r+2}_{n+1,0} \to\bar{E}^{r+1}_{n+1,0}\to E^{r+2}_{n-r,r}\to\bar{E}^{r+2}_{n-r,r}\to0$$

and then $$0\to{\bar{E}}^{r+2}_{n+1,0} \to\bar{E}^{r+1}_{n+1,0}\to E^{\infty}_{n-r,r}\to\bar{E}^{\infty}_{n-r,r}\to0$$

Putting $r=n$ it becomes

$$0\to{\bar{E}}^{\infty}_{n+1,0} \to\bar{E}^{n+1}_{n+1,0}\to E^{\infty}_{0,n+1}\to\bar{E}^{\infty}_{0,n+1}\to0$$

$$H_{n+1}\to \bar{H}_{n+1} \to\bar{E}^{n+1}_{n+1,0}\to E^{\infty}_{0,n+1}\to\bar{E}^{\infty}_{0,n+1}\to0$$

So far so good.

Answer by Wilberd van der Kallen for Subgroups of $GL_n(\mathbb Z)$ with finite coinvariants

2012-07-27T14:19:20Z

For any finite index subgroup $G$ there is a nonzero integer $m$ so that $G$ contains the elementary matrices $e_{ij}(m)$ that have ones on the diagonal, $m$ at the $(i,j)$ entry and zeroes elsewhere. So the span of $\{gz-z: g\in G, z\in\mathbb Z^n\}$ contains all multiples of $m$ and ${\mathbb Z}^n_G$ is finite. Now just take a torsion-free finite index subgroup $G$.

Edit (by Igor Belegradek): I add some detail for my records. If $e_r$ is a vector in the standard basis, then it is easy to write $me_r$ in the form $v-e_{ij}(m)v$ for some $i\neq j$ and $v\in\mathbb Z^n$. It remains to show that for any finite index subgroup $G$ of $GL_n(\mathbb Z)$ all such $e_{ij}(m)$ lie in $G$ for some $m$. By making $G$ smaller we can assume it is normal and of index $k$. Since $e_{ij}(m)=e_{ij}(1)^m$, it suffices to assume that $k$ divides $m$.

Answer by Wilberd van der Kallen for Are the Weyl modules projectives?

2012-06-20T12:15:59Z

There is actually a mildly interesting category in which a given Weyl module is projective. The simplest Weyl module is one dimensional with trivial action. So in that case the category has to have trivial cohomology and is indeed rather uninteresting.

But now consider more generally a Weyl module with highest weight $\lambda$. Choose a Weyl group invariant inner product on the vector space spanned by the weight lattice, so that one can speak of the length of a weight. Then the appropriate category consists of the representations all whose weights have length at most equal to the length of $\lambda$.

This is Polo's theorem as treated (dually) in my Lectures on Frobenius Splittings and B-modules.

But notice that the category depends on the Weyl module.

Answer by Wilberd van der Kallen for Quotient space of algebraic group

2012-06-17T07:50:09Z

You want to show that $T_eH=\{D\in T_eG\mid\delta_DI\subset I\}$. This is something general about smooth subvarieties of a variety. The left hand side maps to the right hand side by functoriality of tangent spaces. Now $H$ is a smooth subvariety and the dimension of its tangent space is the dimension of $H$. So we must understand that the right hand side is no bigger. But for this one may look in local coordinates at $e$. Say $m$ is the ideal of functions vanishing at $e$. Then one may compute with $m/m^2$ to check that preserving $I$ imposes enough linear restrictions to bound the dimension. As an algebraist I would first pass to the $m$-adic completion which is a ring of power series. In that ring the ideal $I$ looks very simple.

The other equality cannot be proved without further data on the construction of $v$. But basically it is again a statement that things are OK in local coordinates, now around $v$. The condition on an element of $T_eG$ that it does not move $v$ must now impose enough linear restrictions to bound the dimension again.

Answer by Wilberd van der Kallen for Can the 'linkages' between equivalent extensions of modules of an algebraic group be taken to have bounded length?

2011-11-24T09:26:39Z

The linkage bound is 2.

If the algebraic group is simple, say over an algebraically closed $k$, then one has the following lemma.

Lemma. If $V$, $W$ are finite dimensional, there is an $m$ depending on $V$, $W$, so that if $St_n$ is the $n$-th Steinberg module with $n\geq m$ the natural map $Ext^i(V,W)\to Ext^i(V,W\otimes St_n\otimes St_n)$ vanishes for $i\geq 1$.

See for instance my 1977 joint paper with Cline, Parshall and Scott. So this gives a way to kill extension classes while staying within the category of finite dimensional representations. Now it seems one may argue as in Sasha's answer, dualized.

Indeed consider a representative $E$ of an element of $Ext^1(V,W)$. Let us write $W\otimes St_n \otimes St_n$ as $I_n(W)$. For $n$ large the lemma implies that $E$ is the Yoneda composite $P\circ f$ of $P:0\to W\to I_n(W)\to I_n(W)/W\to 0$ with an element $f:V\to I_n(W)/W$. Now if one has a representative of $Ext^i(V,W)$ with $i>1$, write it as Yoneda composite $E\circ F$ of an $E\in Ext^1(Z,W)$ and an $F\in Ext^{i-1}(V,Z)$. Applying the previous result one gets a linkage map from $E\circ F$ towards a Yoneda composite $P\circ Q$ of the representative $P:0\to W\to I_n(W)\to I_n(W)/W\to 0$ with a representative $Q=f\circ F$ of an element of $Ext^{i-1}(V, I_n(W)/W)$. So we use that there is a linkage from $(P\circ f)\circ F$ towards $P\circ(f\circ F)$. One repeats this untill one has a linkage map from the original $E\circ F$ towards a Yoneda composite $R\circ S$ of an extension $R:0\to W\to I_{n_1}(W)\to I_{n_2}(X_1)\to \cdots I_{n_{i-2}}(X_{i-2})\to I_{n_{i-2}}(X_{i-2})/X_{i-2}\to 0$ [ depending only on $W$ ] with an element $S$ of $Ext^1(V,I_n(X_{i-2})/X_{i-2})$. Given a second representative of the same class of $Ext^i(V,W)$ one may repeat the construction and if the $n_i$ are big enough [ meaning they grow fast enough ] one ends up with the same class of $Ext^1(V,I_n(X_{i-2})/X_{i-2})$, by a dimension shift argument. So the linkage is bounded by 2 as in Sasha's answer.

For reductive groups over an algebraically closed $k$ the Lemma still holds, interpreted appropriately. For nonreductive groups one may take an exhaustive filtration $k\subset M_1\subset M_2\cdots$ of the coordinate ring $k[G]$ by finite dimensional submodules [ for the action by left translation ] and replace $St_n \otimes St_n$ with $M_n$ in the argument above. The argument needs to be modified a bit. For instance the lemma is no longer available. But the limit over $n$ of the $Ext^i(V,W\otimes M_n)$ vanishes and that is enough to construct all the needed maps.

Answer by Wilberd van der Kallen for A question about the additive group of a finitely generated integral domain

2011-10-10T07:22:42Z

As Qing Liu explains there may be such nontrivial $e$.

Suppose there was such an $e$. By Grothendieck's Generic Freeness Theorem, [Theorem 14.4 in David Eisenbud, Commutative algebra with a view toward algebraic geometry, Graduate Texts in Mathematics, Springer-Verlag, New York, 1995] there is $0\neq a\in \Bbb Z$ so that $A[1/a]$ is a free $\Bbb Z[1/a]$-module. Choose a basis and write $1$ in terms of that basis. We see $1$ lies in a direct summand spanned by finitely many basis vectors. By the structure theorem of finitely generated modules over a PID we see that in fact $Q=A[1/a]/\Bbb Z[1/a]$ is a free $Z[1/a]$-module plus a finite group. So it does not contain any nontrivial divisible element. But the image of $e$ in $Q$ is divisible. That means that $e\in \Bbb Z[1/a]$.

So far so good. The next line is wrong, as explained by Qing Liu.

But $\Bbb Z[1/a]/\Bbb Z$ does not contain any divisible element.

Answer by Wilberd van der Kallen for Isomorphism between varieties of char 0

2011-08-21T12:23:58Z

They must be extremely nice, as the projection of the cusp $y^2=x^3$ onto the $x$-axis is already a counterexample. You want the varieties normal?

Does the action of an affine group scheme preserve the nilradical of an algebra?

2011-06-21T11:59:58Z

Let $k$ be a commutative ring and let $G$ be a flat affine algebraic group scheme over $k$. Let $G$ act by algebra automorphisms on the commutative $k$-algebra $A$. So $G(R)$ acts by $R$-algebra automorphisms on $A\otimes_k R$ for any commutative $k$-algebra $R$. Let $N$ be the nilradical of $A$. Is $N$ always a $G$ submodule? So is the image of $N\otimes_k R$ in $A\otimes_k R$ invariant under $G(R)$? I only need it when $G$ is a Chevalley group scheme, in which case it is true. But my question is if this is a general fact.

Answer by Wilberd van der Kallen for universal cover of SL2(R): does it admit central extensions?

2011-05-01T10:42:00Z

The answer should be negative, because the $K_2$ of the reals is humongous. That is, there are nontrivial central extensions. (Please do not ask a question and then explain its negation.) Algebraic $K$ theory detects transcendentals. There is a Chern class map from $K_2(\Bbb R)$ towards $\Omega^2_{\Bbb R}$, where the Kähler differentials are taken over the integers. It maps the Steinberg symbol {t,u} to $dlog\ t\wedge dlog\ u$, with $dlog\ t=dt/t$. It hits much more than a cyclic group. The map from the Schur multiplier of $SL_2(\Bbb R)$ to the stable $K$ group is surjective in this case, by Steinberg. So the universal cover as a Lie group realizes only a very small part of the Schur multiplier.

Answer by Wilberd van der Kallen for Convergence of eigenvectors

2011-03-25T10:58:27Z

For eigenvectors there is no chance. One may approximate the identity map $T$ on $\Bbb R^2$ with a symmetric matrix $T_n$ whose eigenvalues are $1$ and $1-1/n$. The eigenvectors are perpendicular to each other, but otherwise their direction is entirely optional. So by choosing directions erratically one can avoid convergence. Of course some subsequence will converge.

Answer by Wilberd van der Kallen for Question about the representation theory of SL(n,Z)

2011-03-21T15:49:44Z

Consider the surjective map of $SL(n,\Bbb Z)$-modules $Hom_{\Bbb C}(V',V)\to Hom_{\Bbb C}(V',V')$. Tim tells us that the identity map from $V'$ to $V'$ lifts to an $f:V'\to V$ which is invariant under a finite index subgroup $\Gamma $ of $SL(n,\Bbb Z)$. Then by averageing one can make it invariant under $SL(n,\Bbb Z)$.

Answer by Wilberd van der Kallen for Intersections of conjugates of the icosahedral group in SO(3)

2011-03-21T15:24:18Z

Rotate a quarter turn around the axis passing through the midpoints of two antipodal edges. That gives a different copy of the original icosahedron. A half turn preserves both icosahedra. So the statement is wrong.

Answer by Wilberd van der Kallen for Tensor Product of Witt Vectors

2011-03-04T12:36:32Z

When $B$ is $\acute{\rm e}$tale over $C$ and $A$ or $B$ is finite over $C$, then the result is known by Theorem (2.4) in my paper Descent for the $K$-theory of polynomial rings, link text

Answer by Wilberd van der Kallen for Applications of finite continued fractions

2011-02-20T09:52:45Z

Wim Hesselink posed a problem motivated by image processing of a discretized picture. I found that it was helpful to consider the convergents in a continued fraction approximation of rational numbers. See link text

Answer by Wilberd van der Kallen for Proper subgroup of GL(n,Z) isomorphic to GL(n,Z)?

2010-09-16T09:56:16Z

For $G=GL(2,\mathbb Z)$ there is no proper subgroup isomorphic to it. Consider the dihedral group $D$ of isometries of a regular 6-gon. There is only one conjugacy class in $G$ of subgroups isomorphic to $D$. Indeed given such a subgroup, simply call it $D$, one may adapt the inner product to it, by averaging, so as to make $D$ consist of orthogonal matrices. Then take a basis of $\mathbb Z^2$ consisting of shortest vectors making an obtuse angle, say. Let $s$ be the element that swaps the two basis vectors. We now look for an element $u$ of order four with $susu=1$ so that $u^2$ commutes with the elements of $D$. There is very little choice and we find that $u$ together with $D$ generates $G$.

Answer by Wilberd van der Kallen for Non-finite version of Nakayama's lemma?

2010-07-27T14:32:03Z

Let $N$ be the $A$-module generated by $S$. Now $M$ is contained in $N+\mathfrak{m}M$, which is contained in $N+\mathfrak{m}(N+\mathfrak{m}M)$, hence in $N+\mathfrak{m}^2M$. Repeat.

Answer by Wilberd van der Kallen for Is Lusztig's conjecture solved?

2010-03-28T10:24:31Z

The result of that book is that the conjecture is true for sufficiently large, but unspecified characteristic. (First fix a Dynkin type.) More recently Peter Fiebig has given actual bounds. See

An upper bound on the exceptional characteristics for Lusztig's character formula by Peter Fiebig arXiv:0811.1674v2 at http://arxiv.org/pdf/0811.1674v2

Answer by Wilberd van der Kallen for Connected components of the orthogonal group O(2n) in characteristic 2.

2010-03-25T16:28:44Z

Presumably this is treated in detail in chapter 7 of the book The Classical Groups and K-Theory, by A.J.Hahn and O.T.O'Meara.

On page 424 it says in theorem 7.2.23 that the elementary subgroup has index two. And elementary matrices are in the connected component of 1.

Wilberd

Answer by Wilberd van der Kallen for A ring of invariants in characteristic 2

2010-03-21T16:38:17Z

Indeed the "symmetrized square-free monomials" seem to generate. (Order lexicographically and look what the highest term in a product looks like. Now use that to concoct rewriting rules.)

[Oops! This is less obvious than it seemed. The symmetrizations are not with respect to the full symmetric group. In fact it fails for the cyclic group of order four, where the square-free case is not enough to generate all invariants of degree three.]

They also seem to be independent, as the transcendence degree matches.

[Oops! Also wrong. It would contradict the Chevalley–Shephard–Todd theorem. There may be many orbits of our cyclic group in the set of square free monomials of a given degree.]

One may wish to check the degree of the full ring as a module over the predicted subring. For example, K[x,y] as a module over K[x+y,xy] has basis 1, x, but why?

[Because of the minimal polynomial (T-x)(T-y) over that subring. But this reasoning is less helpful for larger degree. Nevertheless one may wish to look at our full ring as a (free) module over the polynomial ring in the elementary symmetric functions. Is there a basis of that module that is permuted by our cyclic group? And one really wants the ring structure, not just the vector space.]

Wilberd

Comment by Wilberd van der Kallen

2013-06-18T04:04:22Z

If the ideal $I$ is not principal, this makes no difference. Suppose $[\bar f,\bar g]$ is unimodular modulo $I$. Choose $p$, $q\in \Bbb Z[x]$ so that $h:=pf+qg-1\in I$. Then $(f,g,h)$ is a unimodular row that is not reducible.

Comment by Wilberd van der Kallen

2013-04-10T06:57:01Z

@Edward Cooper. Sorry. Try arXiv:1303.60882 by Stepanov.

Comment by Wilberd van der Kallen

2013-04-03T13:26:02Z

You my also need to use Lemma 3.1 of arXiv:1301.6082 to deal with the distinction between the subgroup generated by the $e^\ell_{ij}$ and the normal subgroup generated by them, meaning the smallest normal subgroup of the elementary group that contains them. In the congruence subgroup problem one tends to use normal subgroups.

Comment by Wilberd van der Kallen

2013-03-07T10:01:22Z

That is what it sounded to me when he introduced himself. Definitely not "Munkers".

Comment by Wilberd van der Kallen

2012-12-21T10:58:19Z

@Daniel Litt. You say `This never happens.' That is what Armand Borel wrote. But see Friedrich Knop, Homogeneous varieties for semisimple groups of rank one. Compositio Mathematica, 98 (1995), 77-89. The conclusion is that it is extremely rare, but it does happen.

Comment by Wilberd van der Kallen

2012-10-24T14:53:52Z

@FJH This was not my claim. We wanted to send $v$ to a vector with first component 1, not 0. Zero would also be possible, but that was not the claim. First one adds multiples of later entries to $v_1$ and $v_2$ to arrange that $(v_1,v_2,v_3)$ becomes unimodular. Then one transforms $(v_1,v_2,v_3)$ to $(1,0,0)$. All this leaves the last two coordinates alone and can be achieved by the action of $GL_n$.

Comment by Wilberd van der Kallen

2012-10-24T14:48:15Z

@Andy One should be a little careful about zeroes. When the starting vector is $(8,0,0,0,5)$ it does not suffice to use $r_1$.

Comment by Wilberd van der Kallen

2012-10-24T13:51:05Z

How about taking the average of the position vectors?

Comment by Wilberd van der Kallen

2012-10-09T09:02:26Z

That means using the map $R\to R/J$.

Comment by Wilberd van der Kallen

2012-10-08T12:42:19Z

Well, I was in the setting of Karl, not Proposition 1.5.15 of Bruns and Herzog. Actually B & H tell on the same page that if one wants the degrees unique one should restrict to cases like Karl considers. (They call the degrees $\beta_{0i}$.) But their argument is a bit sophisticated. I was simply observing that if one factors $R$ by the ideal $J$ generated by all elements of degree at least $N$, then the minimal system $S$ maps to a minimal system plus zeroes. Now just watch how many generators become zero as $N$ varies. That tells you how many generators there were in each degree.

Comment by Wilberd van der Kallen

2012-10-08T07:46:30Z

If one applies the positive answer to your original question to truncated versions of the ring, one should get the degrees also. Here by a truncated version I mean that you factor out all elements of degree greater than some fixed number.

Comment by Wilberd van der Kallen

2012-10-08T07:39:43Z

Presumably your generators are homogeneous?

Comment by Wilberd van der Kallen

2012-09-25T09:41:48Z

Because terminology changes over time?

Comment by Wilberd van der Kallen

2012-09-19T15:40:48Z

Also look at $PGL(2,{\bf C})$ viewed as a real Lie group. The complexification of its Lie algebra is not simple.

Comment by Wilberd van der Kallen

2012-09-18T16:05:08Z

A submodule over $A_0$ of $M_n$ can be written as the intersection with $M_n$ of the $A$-module it generates. Now try ascending chains.