Let $V$ be a $2n$-dimensional complex vector space with base $\{e_1,\dotsc,e_n,f_1,\dotsc,f_n]\}$ Let $W \subset \wedge^n V$ be the subspace in the exterior product, with basis vectors $$ e_{i_1} \wedge \dotsb \wedge e_{i_k} \wedge f_{j_1} \wedge \dotsb \wedge f_{j_{n-k}} $$ where we take all possible indices such that $\{i_1,\dotsc,i_k\} \cup \{j_1,\dotsc,j_{n-k}\}$ is a set partition of $\{1,\dotsc,n\}$. Thus, $W$ is $2^n$-dimensional.

As an example, when $n=2$, we have that $W$ has the following four vectors as basis. $$ e_1 \wedge e_2, \quad e_1 \wedge f_2, \quad e_2 \wedge f_1, \quad f_1 \wedge f_2 $$

Suppose now that we have a map $T:V \to V$. It has a natural extension to $\wedge^n V$, (we use $T$ to denote this extension as well) and suppose that $T$ preserves the subspace $W$. Hence, $T$ is also a linear map from $W$ to $W$.

Suppose furthermore that $T$ is diagonalizable, with eigenvalues $x_1,\dotsc,x_{2n}$. Then the trace of the map $T:V\to V$ is simply $x_1+\dotsb+x_{2n}$.

It is straightforward to compute the trace of the induced map $T:\wedge^n V \to \wedge^n V$, it is simply $e_n(x_1,\dotsc,x_{2n})$, where $e_n$ denotes the $n$th elementary symmetric function.

Question I: How can one express the trace of $T:W \to W$? Is the information given even sufficient?

Question II: I am actually only interested in the case when $T:V \to V$ is defined as the cyclic shift, $$ T(e_i) = e_{i+1}, T(e_n)=f_1, T(f_i) = f_{i+1}, T(f_n)=e_1, $$ and powers of $T$. Here, the eigenvalues of $T$ $x_1,\dotsc,x_{2n}$ are simply the roots of $t^{2n}-1=0$.

I think the trace should be $\prod_{j=1}^n (1+\xi^j)$ where $j$ is a primitive $2n$th root of unity, but I cannot really nail down the motivation.

  • $\begingroup$ I have difficulty to understand the definition of $W$ since you wrote it has $2^n$ dimension let us consider $n=2$ it is more convenient to put an order on variables $x_1<x_2<y_1<y_2$ so I think you mean a base for $W$ is $x_1 \wedge y_2, x-2\wedge y_1 $ so it is 2 dimensional not 4 dimensional space. what is my error? $\endgroup$ – Ali Taghavi Aug 21 '19 at 20:37
  • 1
    $\begingroup$ @AliTaghavi By convention, there is an order of the indices - increasing. Note that if you know the e-indices, then the f-indices are known as well. The subset of [n] that index the e-part can be chosen in 2^n ways. $\endgroup$ – Per Alexandersson Aug 21 '19 at 20:41
  • $\begingroup$ Thanks for your edit and giving an example clearing the definition. $\endgroup$ – Ali Taghavi Aug 21 '19 at 20:52
  • $\begingroup$ is it true (and obvious) that every every linear symplectomorphism of $\mathbb{R}^{2n}$ preserve $W$? $\endgroup$ – Ali Taghavi Aug 21 '19 at 20:54
  • $\begingroup$ Is there an interesting NONLINEAR analogy for the spaces you are considering? $\endgroup$ – Ali Taghavi Aug 21 '19 at 20:56

Q1 There is no answer as it depends not only on $T$ but on its interaction with the decomposition of $V$. For instance, if $n=2$, $T(e_1)=T(e_2)=0, T(f_i)=f_i$ and $T'(e_1)=T'(f_1)=0, T'(f_2)=f_2, T'(e_2)=e_2$ are the same as linear operators, but their restrictions to $W$ are different.

Q2 The answer is $0$.

Your operator on $W$ is monomial. Just make sure that no standard basis element goes into a multiple of itself.

  • 1
    $\begingroup$ And of course, the OP's guess $\prod_{j=1}^n \left(1+\xi^j\right) = 0$ as well, due to the $1 + \xi^n = 1 + \left(-1\right) = 0$ factor :) $\endgroup$ – darij grinberg Aug 22 '19 at 17:31
  • $\begingroup$ @BugsBunny How about powers of T then? $\endgroup$ – Per Alexandersson Aug 22 '19 at 17:36
  • $\begingroup$ @Per Alexandersson It is doable as well. It is equivalent to figuring out all eigenvalues. It is clear how to do it for each $n$ but I am too drunk to think of a general formula. The basis elements correspond to 2-ary length $n$ necklaces (see en.wikipedia.org/wiki/Necklace_(combinatorics) ). Thus, you need to know the sizes of all the necklaces. Each necklace of size $m$ contributes all roots of $z^m-(-1)^m$ to the eigenvalues. Now it is up to your enumerative combinatorics skills (mine are $-\infty$) to figure out the general answer. $\endgroup$ – Bugs Bunny Aug 22 '19 at 18:12
  • $\begingroup$ @BugsBunny Ah, i see. Yes, i have already done the enumerative part - I was hoping for a way to avoid that in the proof. I am trying to prove a cyclic sieving phenomenon by using representation-theory, see background here: math.upenn.edu/~peal/polynomials/… $\endgroup$ – Per Alexandersson Aug 23 '19 at 5:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.