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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.

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  • $\begingroup$ 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, 2019 at 20:37
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    $\begingroup$ @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, 2019 at 20:41
  • $\begingroup$ Thanks for your edit and giving an example clearing the definition. $\endgroup$
    – Ali Taghavi
    Aug 21, 2019 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, 2019 at 20:54
  • $\begingroup$ Is there an interesting NONLINEAR analogy for the spaces you are considering? $\endgroup$
    – Ali Taghavi
    Aug 21, 2019 at 20:56

1 Answer 1

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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.

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    $\begingroup$ 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, 2019 at 17:31
  • $\begingroup$ @BugsBunny How about powers of T then? $\endgroup$
    – Per Alexandersson
    Aug 22, 2019 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, 2019 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, 2019 at 5:09

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