User jesko hüttenhain - MathOverflow

A basis of the symmetric power consisting of powers

2013-04-17T08:44:41Z

I have asked this question on math.se, but did not get an answer - I was quite surprised because I thought that lots of people must have though about this before:

Let $V$ be a complex vector space with basis $x_1,\ldots,x_n\in V$. Denote by $v_1\odot\cdots\odot v_k$ the image of $v_1\otimes\cdots\otimes v_k$ in the symmetric power $\newcommand{\Sym}{\mathrm{Sym}}\Sym^k(V)$. It is well-known that the Elements $v^{\odot k}$ for $v\in V$ generate this space (see, for instance, this answer on math.se), so they must contain a basis.

In other words, let $N=\binom{n+k-1}k$, then there must be $v_1,\ldots,v_N\in V$ with $$\mathrm{Sym}^k V = \mathbb Cv_1^{\odot k} \oplus \cdots \oplus \mathbb C v_N^{\odot k}.$$ I am looking for an explicit description of such a basis. Is such a description known? Is there maybe even a "nice" or somewhat "natural" choice for the $v_i$ as linear combinations of the $x_i$?

Expression of basis vectors of permutation modules in different bases.

2013-05-08T13:01:42Z

This is a cross-post from math.se, because I did not get any answer there:

Write $[n]:=\{1,\ldots,n\}$. For a partition $\lambda\vdash n$, I will write $[\lambda]$ for the Specht module that corresponds to $\lambda$, i.e. the complex vector space spanned by all standard tableaux of shape $\lambda$.

Consider the complex vector space $V_k$ which has as a basis the $k$-element subsets of $[n]$. The symmetric group $S_n$ acts on $V_k$ in the was that $g\in S_n$ maps a basis vector $A\subseteq[n]$ to the basis vector $g(A)=\{ g(i) \mid i\in A \}$. Let us assume $k\le \frac n2$. It is known that $$V_k = \bigoplus_{j=0}^k~ [n-j,j]$$ is the decomposition of $V_k$ into irreducibles. See, for instance, Example 7.18.8 in Stanley's book Enumerative Combinatorics (Volume 2).

Question: Given any $j\le\frac n2$ and a standard tableau $T$ of shape $\lambda=(n-j,j)$, I would like to know how to express the corresponding basis vector $v_T$ of $[\lambda]$ in the canonical basis of $k$-element subsets of $[n]$.

Thoughts: For $k=1$, this is rather easy. For the only standard tableau $T$ on $\lambda=(n)$, the vector $v_T=\sum_{i=1}^n \{ i \}$ is just the sum of all $1$-element subsets of $[n]$. On the other hand, for a standard tableau $T$ of $(n-1,1)$, let $a_T$ be the number in the single box in row $2$. Then it can be checked that $v_T=\{1\}-\{a_T\}$. However, I already have trouble with $V_2$. There is an obvious $S_n$-invariant inclusion \begin{align*} \phi_1: V_1 &\longrightarrow V_2 \\ \{i\} &\longmapsto \sum_{j\ne i} \{ i,j \} \end{align*} and the $[n-2,2]\subseteq V_2$ must be isomorphic to $V_2/\mathrm{im}(\phi_1)$, but I didn't get much further.

The notion of multiplicity in algebraic geometry

2013-04-22T19:52:38Z

Let $A$ be a commutative ring. Let $f\in A\setminus\{0\}$ and $I\subseteq A$ any ideal. I would like to define the multiplicity of $f$ at $I$ as $$\mu_f(I):= \max\{\, d\ge 0 \mid f\in I^d\,\},$$ where $I^0:= A$ . In the case where $A$ is Noetherian and either local or an integral domain, the Krull Intersection Theorem (Eisenbud, Corollary 5.4) implies that $\mu_f(I)$ is well-defined. Main scenario: $A$ is the local ring of a locally Noetherian scheme $X$ at some point $P$ , $I$ is the corresponding maximal ideal and $f$ is locally representing a Cartier divisor on $X$ .

I have only seen this in Hartshorne, Page 388, for surfaces, but I do not see why the definition should be limited to surfaces. In general, I only know the following definition of geometric multiplicity, for locally Noetherian schemes $X$ and points $P\in X$ of codimension one: $$\bar\mu_f(P):=\mathrm{length}_{\mathcal O_{X,P}}(\mathcal O_{X,P}/(f))$$ Does this coincide with the above definition? If yes, why is $\bar\mu$ so prominent? After all, $\mu$ is more general.

Syzygies of determinantal varieties: Looking for English text

2013-04-10T17:32:31Z

I would like to understand the syzygies of the determinantal ideal $I_r$, generated by the $r\times r$ minors of a matrix $(X_{ij})$ of indeterminantes in the polynomial ring over an algebraically closed field of characteristic zero. The original resource for this object of study is the paper "Syzygies des variétés déterminantales" by Alain Lascoux [L]. While I would like to read it at some point, my rusty French is making it a bit cumbersome, and hence I was wondering if there were any translations of this treatment in English, possibly in some textbook. Thanks a lot in advance already.

[L] A. Lascoux, Syzygies des variétés déterminantales, Adv. Math. 30 (1978), 202–237.
Mathematical Reviews (MathSciNet): MR520233
Digital Object Identifier: doi:10.1016/0001-8708(78)90037-3

Answer by Jesko Hüttenhain for Syzygies of determinantal varieties: Looking for English text

2013-04-10T17:40:23Z

A less sophisticated search was more successful. It's all there in Chapter 6 of the book Cohomology of Vector Bundles and Syzygies by Weyman.

Since I can't accept my own answer for another two days, feel free to post any other suggestions, though. More reading material can't hurt.

The shortest mathematical paper

2013-04-10T09:15:10Z

I was looking at the paper Zum Hilbertschen Nullstellensatz [1] and wondered if there was a shorter mathematical paper than this one. A colleague of mine rumored about a number-theoretic paper where just a counterexample consisting of a few numbers is given, and he conjectured that it must be the shortest mathematical paper. However, we were unable to find it. I would be very curious to see it.

So, I am asking you for references to papers that are shorter at least than [1] and if possible, as short as said number-theoretic paper. Thanks a lot in advance. Also, I am sorry if this question is too far off-topic.

[1] J. L. Rabinowitsch, Zum Hilbertschen Nullstellensatz, Mathematische Annalen Volume 102, Number 1 (1929), 520.

The Hilbert function of an intersection

2013-04-04T18:23:04Z

Assume that $X_1,\ldots,X_r\subseteq\mathbb P^n$ are irreducible, reduced hypersurfaces in complex projective space, each of the same degree $d$. In other words, $X_i=Z_\ast(f_i)$ for certain irreducible, homogeneous polynomials $f_i\in\mathbb C[X_0,\ldots,X_n]_d$. Let $X:=X_1\cap\cdots\cap X_r$ be their intersection. For the moment, let's just say scheme-theoretic intersection. If it helps, assume that the codimension of $X$ is $r$, i.e. it is the complete intersection of the $X_i$. I would like to know what is known, if anything, about the Hilbert function of the closed subscheme $X$, or equivalently, the ideal $I=(f_1,\ldots,f_r)$. Can it be expressed in terms of the $f_i$ somehow?

Not very important, but still interesting: What about passing to $\sqrt I$?

Decomposition of $\mathrm{End}(V)$ as $S_n\times S_n$-module

2013-02-28T13:29:34Z

Let $V$ be a finite-dimensional, complex vector space and set $\newcommand{\Gl}{\mathrm{Gl}}G:=\Gl(V)\times\Gl(V)$. Let $E:=\mathrm{End}(V)$ and consider its coordinate ring $\mathbb C[E]$, the space of all polynomial functions on $E$. It is well-known (see 9.7 in Claudio Procesi's book on Lie Groups) that as a $G$-module, $\mathbb C[E]$ decomposes as \begin{align*} \mathbb C[E]_d &= \bigoplus_{\lambda\mathrel\vdash d} \mathbb S_\lambda(V^\ast)\otimes\mathbb S_\lambda(V) \end{align*} where $\mathbb S_\lambda$ denotes the Schur functor. Now, simply by choosing a basis and restricting to permutation matrices, we have an action of $S:=S_n\times S_n$ on $E$ and therefore, also on $\mathbb C[E]$. Hence, there must be some decomposition \begin{align*} \mathbb C[E]_d &= \bigoplus_{\lambda,\mu \mathrel\vdash n} (\mathbb V_\lambda\otimes \mathbb V_\mu)^{\oplus N_d(\lambda,\mu)} \end{align*} where $\mathbb V_\lambda$ is the Specht module and $N_d(\lambda,\mu)\in\mathbb N$ are certain multiplicities.

My question is: What, if anything, is known about the $N_d(\lambda,\mu)$?

Question about local description of the branch locus

2013-01-29T12:21:49Z

Let $\pi:Y\to X$ be a dominant, finite morphism of nonsingular varieties over an algebraically closed field $\Bbbk$ . Assume furthermore that for all $Q\in Y$ , with $P=\pi(Q)$ , we have $$\mathcal O_{Y,Q}=\mathcal O_{X,P}[T_1,\ldots,T_k]/(T_1^n-x_1,\ldots,T_k^n-x_k)$$ for certain $x_i\in\mathcal O_{X,P}$ . In other words, we have adjoined certain $n$-th roots.

Now, let $R\subseteq Y$ be the ramification divisor and $H=\pi(R)$ the branch locus (both now with the reduced subscheme structure). Let $\mathcal I(H)$ be the ideal sheaf of $H$ and $\mathcal I(H)_P$ the stalk at $P$ . Can I conclude that this ideal $\mathcal I(H)_P\subseteq \mathcal O_{X,P}$ is generated by the product $x_1\cdots x_k$ ? Or, correspondingly, can I conclude that $\mathcal I(R)_Q$ is generated by the product $y_1\cdots y_k$ ?

Edit. As pointed out in the answer by Dmitry Vaintrob, I want to assume $n$ not divisible by $\mathrm{char}(\Bbbk)$ . Furthermore, we assume that the $x_i$ are reduced and coprime.

Top chern class under finite, unramified, dominant morphism

2013-01-20T11:49:57Z

Situation: Let $\Bbbk$ be an algebraically closed field. Assume that $\pi:Y\to X$ is an finite, dominant, unramified morphism between nonsingular varieties of dimensions $n$. Let $d=\deg(\pi)$.

What I know: For $\Bbbk=\mathbb C$, the second chern class of $X$ equals its topological Euler characteristic (i.e., the Euler characteristic with respect to the topology of a complex manifold). I know this under the name Gauss-Bonnet Formula. It then follows that $c_n(Y)=d\cdot c_n(X)$ because $\pi$ is a $d$-fold covering map of complex manifolds.

My Question: Does $c_n(Y)=d\cdot c_n(X)$ hold under the more general assumption that $\Bbbk$ is algebraically closed? In particular, does this hold in positive characteristic?

PS: I am mostly interested in the case $n=2$, i.e. $\pi$ is a covering map of surfaces. However, I felt that this would probably work for any $n$.

Higher dimensional version of the Hurwitz formula?

2011-08-03T01:53:51Z

In Hartshorne IV.2, notions related to ramification and branching are introduced, but only for curves. The main result is the Hurwitz formula.

Now if you have a finite surjective morphism between nonsingular, quasi-projective varieties, then the notion of ramification (divisor) would still make sense and we can also still talk about the degree of a canonical divisor. It also seemed to me like no result in IV.2 really uses the fact that $X$ and $Y$ are of dimension $1$. So I ask, can I replace $f$ by a finite, dominant, separable morphism $X\to Y$ of nonsingular, quasi-projective varieties of arbitrary dimension? That is, of course, up to and including Proposition 2.3.

If this is so, can we say anything about the degree of a canonical divisor in dimension greater than one? Maybe in special cases?

Top chern class in positive characteristic

2011-11-11T17:17:20Z

Given a nonsingular, projective variety $X$ of dimension $n$ over an algebraically closed field $k$. Over $k=\mathbb{C}$, the top chern class $c_n(T_X)$ of the tangent sheaf is the Euler characteristic of the associated complex manifold. Is there some kind of (geometric) intuition or well-known formulas for the value of $c_n(T_X)$ in the case of $p=\mathrm{char}(k)>0$?

Intersection theory for $G$-varieties - an action on the chow ring?

2012-10-17T16:05:10Z

Let $G$ be a reductive algebraic group. Let $X$ be a $G$-variety and consider any closed subvariety $Z$ of $X$. Since any $g\in G$ acts as an automorphism, we know that $g.Z$ is again a closed subvariety of $X$. This yields an action of $G$ on the free module of cycles of $X$ which should induce an action of $G$ on the Chow ring of $X$. The invariants of this ring should be precisely the classes that correspond to linear combinations of $G$-orbits.

Has this action been studied before? Any kind of reference would be very welcome. Thanks!

Edit: It looks like my above idea is rather futile, so let me ask more broadly: Are there any techniques or results in intersection theory specifically on $G$-varieties? Could you name some references?

When is an orbit spherical?

2012-09-07T09:03:53Z

I asked the following question over at math.stackexchange, but got no answers. Maybe it's less well-known than I thought, but I still wanted to ask here:

Let's assume we have an affine, reductive, algebraic group $G$ acting algebraically on a variety $X$, everything over an algebraically closed field of characteristic zero. Let $x\in X$ be some point with reductive stabilizer $H:=G_x$. Under what conditions on $x$ or $H$ is the orbit $G.x\cong G\newcommand{\qq}{/\hspace{-.8ex}/}\qq H$ a spherical variety? Let me briefly recall that a spherical variety is a homogeneous space $G\qq H$ satisfying one of the following, equivalent properties:

Any Borel subgroup $B\subseteq G$ has an open orbit in $G\qq H$.
Every equivariant completion of $G\qq H$ contains only finitely many orbits.
For every irreducible $G$-module $V$ and any character $\chi$ of $H$, $$\dim\left\{~v\in V \mid \forall h\in H: h.v = \chi(h)v ~\right\}\le 1.$$

I was hoping that this is well-known, but I cannot find any direct statements of that kind. Searching for the keywords "orbit" and "spherical" is quite fruitless because of property 1.

Edit: In the cases of interest to me, the orbit $G.x$ is affine.

About the strength of representation-theoretic obstructions for orbit closure problems

2012-09-03T16:18:42Z

Let $G$ be a reductive, affine, algebraic group over $\newcommand{\C}{\mathbb C}\C$ . Let $X$ be a $G$ -variety. For $x\in X$ , we write $$G_x:=\{ g\in G\mid g.x=x\}$$ for its stabilizer and for any subgroup $H\subseteq G$ , we write $$X^H:=\{x\in X\mid H.x=x\}$$ for the $H$ -invariants of $X$ . We say that $x\in X$ is characterized by its stabilizer if $X^{G_x}=\{x\}$ . Let $\{V_\lambda\mid \lambda\in\Lambda\}$ be the irreducible $G$ -modules.

Given two points $x,y\in X$ , then $x\in\overline{G.y}$ implies $\overline{G.x}\subseteq\overline{G.y}$ . Hence, $\C[\overline{G.y}]\twoheadrightarrow\C[\overline{G.x}]$ and thus,

$$\DeclareMathOperator{\mult}{mult}\forall \lambda\in\Lambda:\quad \mult\nolimits_\lambda(\C[\overline{G.x}])\le\mult\nolimits_\lambda(\C[\overline{G.y}])$$

Finding $\lambda\in\Lambda$ violating the above is therefore an "obstruction" for the inclusion of orbit closures.

My question now is the following: If $x$ and $y$ are characterized by their respective stabilizers, does the converse hold? I.e., does the above inequality imply that $x\in\overline{G.y}$ ? I have been trying to come up with a counterexample, but without success so far.

Intuition: If $G$ acts on a variety $Y$ and $y\in Y$ is characterized by its stabilizer, then you can very easily find counterexamples if you give up the condition that both points are characterized by their respective stabilizers: Consider $X:=Y\times\{z_1,z_2\}$ with $G$ acting trivially on $Z=\{z_1,z_2\}$ . Now, the points $x_i:=(y,z_i)$ satisfy $x_1\notin\overline{G.x_2}$ and $\C[\overline{G.x_1}]\cong\C[\overline{G.x_2}]$ . In the cases of interest to me, however, both points are characterized by their stabilizer and the question arises whether there are counterexamples under this additional condition.

Action of k* on a variety induces grading?

2012-08-15T10:21:25Z

Let $V$ be a $\Bbbk$-variety such that $\Bbbk^\times$ (as an algebraic group) acts algebraically on $V$. Given any $f\in\Bbbk[V]$, let us call $f$ homogeneous of degree $d$ if for all $v\in V$ and all $\lambda\in\Bbbk^\times$, we have $f(\lambda.v)=\lambda^d f(v)$.

My question is: Does this define a grading on $\Bbbk[V]$?

I was convinced that it is true, but I am running into difficulties. Let us first assume $\Bbbk=\mathbb{C}$, the ground field should not be an obstruction. The linear span of $\Bbbk^\times f$ decomposes since $\Bbbk^\times$ is reductive, but I don't see how to turn this into a grading on all of $\Bbbk[V]$.

If it is true, I would really like to see a proof - it should use as little machinery as possible.

Answer by Jesko Hüttenhain for Wonderful applications of the Vandermonde determinant

2012-07-31T21:17:21Z

Especially for students with just very basic background, it might be a fun fact that companion matrices are diagonalized by the vandermonde matrix corresponding to the zeros of the characteristic polynomial they encode - provided, of course, that the roots are all distinct.

Answer by Jesko Hüttenhain for Minimal degree of polynomial vanishing on the variety of small degree.

2012-07-30T16:47:13Z

The minimal degree of a polynomial that vanishes on $V$ is the minimal $m$ such that $h_I(m)\ne 0$, where $h_I$ denotes the Hilbert function of the variety $V$. Hence, you are interested in upper (and lower?) bounds on the Hilbert function of certain varieties. There are two papers I know that deal with such bounds:

Answer by Jesko Hüttenhain for Question about Terminology in Mumford

2012-07-26T07:33:31Z

The join of two varieties $X,Y\subseteq \mathbb{P}^n$ is $$ J(X,Y) = \overline{\bigcup_{\substack{x\in X,~y\in Y\\x\ne y}} \ell(x,y)}$$ where $\ell(x,y)$ denotes the projective line through $x$ and $y$. The join of $k$ varieties $X_1,\ldots,X_k\subseteq \mathbb{P}^n$ is defined to be the closure of the union of the corresponding, projective $(k-1)$-folds, or by induction $$J(X_1,\ldots,X_k) := J(X_1,J(X_2,\ldots,X_k))$$

This definition is from Joseph Landsberg's book Tensors: Geometry and Applications, page 118. The graph of a regular function is a projective variety, so this should be defined.

Answer by Jesko Hüttenhain for Proofs that require fundamentally new ways of thinking

2012-07-26T07:26:36Z

I am surprised that noone mentioned Hilbert's proof of Hilbert's Basis Theorem yet. It says that every ideal in $\mathbb{C}[x_1,\ldots,x_n]$ is finitely generated - the proof is nonconstructive in the sense that it does not give an explicit set of generators of an ideal. When P. Gordan (a leading algebraists at that time) first saw Hilbert's proof, he said, "This is not Mathematics, but theology!"

However, in 1899, Gordan published a simplified proof of Hilbert's theorem and commented with "I have convinced myself that theology also has its advantages."

Box-dual of a partition - what is it called?

2012-05-24T07:12:17Z

Fix natural numbers $n,m\in\mathbb{N}$. Given a partition $\lambda\vdash d$ with at most $n$ rows (and at most $m$ columns), we can define a partition $\lambda^\ast=(\lambda^\ast_1,\ldots,\lambda^\ast_n)$ by setting $\lambda^\ast_i:=m-\lambda_{n+1-i}$. Graphically, it is obtained by taking the complement of the Young diagram of $\lambda$ inside the $n\times m$ $-$ box $(m^n)$. Now, I'd be curious to learn about any combinatorial results involving this construction - but I don't know what it is called. If someone could tell me, that would be great - this way I'd not have such a hard time searching.

Answer by Jesko Hüttenhain for Box-dual of a partition - what is it called?

2012-07-23T16:14:18Z

The "right" answer was given by Gjergji Zaimi in his comment above: Both "reverse partition" and "complementary partition" are terms that seem to be in use.

Closure of an orbit under the action of an algebraic group

2012-07-18T09:10:26Z

Setting:

Fix some field $k$. I am not very prudent about the field - although I'd prefer to assume as little as possible, you may assume as much as you want, the case of primary interest being $k=\mathbb{C}$.

Let $G\subseteq\mathrm{Gl}_n(k)$ be a closed, reductive subgroup and $X\subseteq\mathbb{A}_k^m$ an affine $G$-variety. Assume that both $G$ and $X$ are cones, i.e. cut out by homogeneous equations. Note that $\mathrm{Gl}_n(k)$ is the open affine subset of $\mathbb{A}^{n\times n}_k$ where the determinant (a homogeneous polynomial) does not vanish. This yields an action of $k^\times$ on both $G$ and $X$. Under this action, assume that the action map $$\begin{align*}\alpha_x:G&\longrightarrow X \\ g&\longmapsto g.x\end{align*} $$ is a morphism of $k^\times$-varieties for each $x\in X$, i.e. $\lambda g.x=g.\lambda x$ for $g\in G$ and $\lambda\in k^\times$.

Question:

I have a point $x\in X$ such that $H:=G_x$ is reductive. Let $U:= G.x$ be the orbit of $x$. Now in this very friendly setting, I have some questions about the closure $Z:=\overline U$.

How does the coordinate ring $k[Z]$ of $Z$ look like? It is known that $U$ itself is affine, and its coordinate ring can be described as the $H$-invariants of $k[G]$. However, what about $k[Z]$?
The orbit $U$ is smooth, but $Z$ isn't (in general). Since $Z$ is the union of orbits, however, the singular locus of $Z$ should also be a union of orbits. Is there some nice way to characterize $\mathrm{Sing}(Z)$?

A particular Isomorphism of graded algebras over a regular local ring

2011-11-02T11:16:26Z

In Hartshorne's "Algebraic Geometry", the following statement is a weaker form of Theorem 8.21A (e), which he quotes from Matsumuura's book on commutative algebra:

Proposition. Let $R$ be a regular local ring and $I=(x_1,\ldots,x_r)\subset R$ an ideal generated by a regular sequence. Let $A:=R/I$. Then, $$\begin{eqnarray*} \phi: A[T_1,\ldots,T_r] &\overset{_\sim}{\longrightarrow} & \mathrm{gr}_I(R) = \bigoplus\nolimits_{d\ge 0} {I^d}/{I^{d+1}} \\ T_i & \longmapsto & x_i \end{eqnarray*}$$ is an isomorphism of graded $A$-algebras.

In Hartshorne, the condition of being regular and local is strengthened to Cohen-Macaulay. However, I only need the above. I tried to look up the proof for the general statement in Matsumuura's book, but it seems rather involved (and honestly, a bit convoluted). I would like to use results about regular local rings and in turn, avoid introducing terminology like Hilbert-Samuel polynomials.

So, I guess I am asking for an "easy" proof of the above proposition. It seems rather easy for $r=1$, but I am somehow stuck trying to prove it by induction.

A polynomial homomorphism from Gl to the group of units is a power of the determinant

2012-07-13T06:39:28Z

I was browsing MO and stumbled upon this post, and I got very curious. I searched for about half an hour and could not find a proof for the statement that any polynomial group homomorphism $\mathrm{Gl}_n(\Bbbk)\to\Bbbk^\times$ is a power of the determinant. Now this is certainly classic, so could someone point me to a good reference? I would prefer a proof that can be presented to graduate students with a mild background in classical algebraic geometry.

Edit: I was also wondering about the following: If $G$ is an affine algebraic group (i.e. an affine $\Bbbk$-variety with compatible group structure), then there is a closed immersion of $\iota:G\hookrightarrow\mathrm{Gl}_n(\Bbbk)$. From such an immersion, I get a "determinant" on $G$, namely $\iota^\sharp(\det)$. Is this, by any chance, independent of the embedding up to taking powers?

What are "maps" between proper classes?

2011-04-28T07:35:23Z

When defining a functor (between categories), I am usually told that it assigns to each object of the source category an object of the target category. I do not find this very satisfactory since we are dealing with proper classes here. Judging by the definition, it must be possible to have the concept of a "map" between proper classes. I would like to know what exactly that is and how it is defined.

I have attempted to read some books on set theory in search for an answer, but they all treat classes very briefly and never mention the possibility of having anything like a map between two of them. I would be just as happy if you could point me to a book where this is explained.

Answer by Jesko Hüttenhain for Pencil of lines and degree $d$ curve in $\mathbb{CP}^2$

2012-05-24T05:59:59Z

Let $\tilde\pi:\mathbb{PC}^2\setminus\{P\}\to\mathbb{PC}$ be the projection from $P$, which restricts to a surjective morphism $\pi:C\to\mathbb{PC}$ of degree $d$. You are asking for the number of points of ramification of this morphism with multiplicity, or more precisely the degree of the ramification divisor $R$. By Hurwitz' theorem, it can be computed as $$\deg(R) = 2g(C)-2 - d(2g(\mathbb{PC}) - 2)=2\binom{d-1}{2} + 2(d-1) = d(d-1).$$

Question about decomposition of exterior product

2012-05-17T09:02:23Z

In their paper "New lower bounds for the border rank of matrix multiplication", Landsberg and Ottaviani make use of the fact that

$$\tag{$\dagger$} {\textstyle\bigwedge}^p(V\otimes W) \cong \bigoplus\nolimits_{\substack{\lambda\vdash p\\\\\ell(\lambda)\le n\\\\\lambda_1\le m }} \mathbb{S}_\lambda V \otimes \mathbb{S}_{\bar\lambda}W$$

where $\bar\lambda$ denotes the conjugate partition of $\lambda$. This isomorphism is basically Exercise 6.11 in Fulton & Harris, so there is no doubt about it. However, from what I gather, in Lemma 3.1 of the paper, they use the fact that the above isomorphism is given by the map

$$ (v_1\otimes w_1)\wedge\ldots\wedge(v_p\otimes w_p) \longmapsto \sum\nolimits_{\substack{\lambda\vdash p\\\ell(\lambda)\le n\\\lambda_1\le m }} c_\lambda(v_1\otimes\ldots\otimes v_p) \otimes c_{\bar\lambda}(w_1\otimes\ldots\otimes w_p), $$

where $c_\lambda$ denotes the Young symmetrizer corresponding to the partition $\lambda$. I cannot find a proof for this. Can someone explain to me why the above map defines

a) a morphism of $\mathfrak{S}_p$-modules and
b) a bijection?

Since all vector spaces involved are of finite dimension and by $(\dagger)$, it would certainly suffice to show that it is either injective or surjective.

Also, if I misunderstood the proof of Lemma 3.1 and the isomorphism is given by another elementary rule, please tell me what it is.

Total exterior Product

2012-05-03T14:19:09Z

In his paper Gaussian maps and plethysm, Manivel uses the term "total exterior product" of line bundles, appearing for the first time on page 3. Given two (projective) varieties $V_1$ and $V_2$ and sheaves of $\mathcal{O}_{V_i}$-modules $\mathcal{F}_i$ (possibly line bundles), there should be some way to construct a sheaf $\mathcal{F}$ on $V_1\times V_2$ which we call the total exterior product of $\mathcal{F}_1$ and $\mathcal{F}_2$. However, I was unable to find a formal definition anywhere in the literature. So, my question is, what is the total exterior product?

Why can I divide an affine variety by the action of the general linear group?

2012-04-17T16:44:22Z

Let $G\subseteq\mathrm{Gl}_n(\mathbb{C})$ be a subgroup of the general linear group and assume that $\rho:G\to\mathrm{Gl}(V)$ is a representation. Understand the complex vector space $V$ as an affine algebraic variety. Then, it appears to be well-known that the quotient $V/G$ has the structure of an algebraic variety such that the quotient map $\pi:V\to V/G$ is a morphism of varieties. However, I cannot find a proof for this statement. There is an abundance of proofs for the case where $G$ is finite, using the Reynolds operator and corollaries of Hilbert's basis theorem, but I would like to see a proof in the general case. Thanks very much in advance!

Edit: I forgot to mention that I assume $G$ to be reductive and the action on $V$ is regular. You are free to assume even more about $G$ if that allows you to provide a comprehensible reference for a proof.

Comment by Jesko Hüttenhain

2013-05-08T08:18:26Z

Hey! This sounds like a really great approach, but I must confess I do not see (at all) how the formula in Theorem 9.6.1 implies the claim that all the $(\sum_i \alpha_i x_i)^k$ form a basis of the degree-$k$ polynomials. Would you mind giving a little more detail? I would really appreciate it.

Comment by Jesko Hüttenhain

2013-04-24T13:00:50Z

Thanks a bunch. I accepted this answer because it helps me most for what I am doing, but thanks to Filippo Edoardo and Will Savin for the very helpful explanations.

Comment by Jesko Hüttenhain

2013-04-23T18:42:34Z

Your definition of $\mathrm{ord}(f)$ confuses me because for one thing, it does not depend on $f$ at all and second, $I\in m^i$ looks a lot like you ment to write $f\in m^i$.

Comment by Jesko Hüttenhain

2013-04-23T07:22:54Z

I suppose $\mathrm{ord}(f)=\sup\{ i\mid f\in m^i \}$ , yes? So basically, I can use my definition in the case where $X$ is a regular scheme, which might just be good enough for me. +1 and thanks!

Comment by Jesko Hüttenhain

2013-04-23T07:14:38Z

Not that it really matters, but I thought $3$ sounded reasonable: We have the ring $k[t^2,t^3]/(t^4)=k[x,y]/(x^2,x^3-y^2)$ where $x=t^2$, $y=t^3$ and $xy=t^5$. One more question, though: You say that the Definitions agree when the local ring is a DVR: Do you have a reference?

Comment by Jesko Hüttenhain

2013-04-22T20:45:36Z

Yea, that was exactly my suspicion. I would like to know how (or rather when) it misbehaves, though - most of the time I am dealing with very forgiving kinds of schemes anyway.

Comment by Jesko Hüttenhain

2013-04-17T12:08:27Z

Oh, don't get me wrong, these elements were my first choice as well and I very much believe that they work, but I can't prove it.

Comment by Jesko Hüttenhain

2013-04-17T09:44:19Z

That's basically the elements from the proof I referenced, but for $k>2$ it requires more than $\binom{n+k-1}k$ terms, namely all the $(x_{i_1}+\cdots+x_{i_j})^k$ for $2\le j\le k$. It's not obvious to me why only the ones for $j=k$ should suffice.

Comment by Jesko Hüttenhain

2013-04-11T16:21:14Z

Thanks a lot, this does indeed look very helpful.

Comment by Jesko Hüttenhain

2013-04-10T09:39:26Z

@Michael: Done and thanks, that thread's got some great examples.

Comment by Jesko Hüttenhain

2013-03-01T08:51:18Z

It is a start, indeed, but I don't really know how to compute $\langle s_\lambda, s_\mu[h]\rangle$ either. In fact, I haven't even fully understood Stanley's definition of plethysm, but maybe I have a better chance after reading the Appendix from the start.

Comment by Jesko Hüttenhain

2013-02-10T11:39:49Z

A personal matter came up and I missed the deadline for assigning the bounty, though I certainly would have done so. I even sent a mail to the moderators, but apparently this mistake is irreversible. I hope my heartfelt thanks are enough then =).

Comment by Jesko Hüttenhain

2013-02-05T08:13:45Z

This is exactly what I am asking. I should have added that I assume $n$ not divisible by $\mathrm{char}(\Bbbk)$ and of course, yea, I assume the $x_i$ reduced and coprime. I was pretty sure that this is true, but I wanted confirmation. To be honest, I don't perfectly understand your argument, though: Can I somehow argue that $Y$ locally looks like a fiber product? What exactly do you mean by $nT_1^{n-1}=0$, you kinda took the derivative there? I'm sorry, I just don't perfectly understand. Anyhow, it's precisely what I want and if you can elaborate, you very much deserve the bounty =).

Comment by Jesko Hüttenhain

2013-01-22T17:58:10Z

PS: I ask because that corollary is, in fact, of serious interest to me =).

Comment by Jesko Hüttenhain

2013-01-22T17:57:39Z

First of all, +1 and thanks for the very detailed Answer. I only know Mayer-Vietoris for singular Homology, is there an equivalent for the $\ell$-adic one?