# Trace and exterior product

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.

• 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? – Ali Taghavi Aug 21 '19 at 20:37
• @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. – Per Alexandersson Aug 21 '19 at 20:41
• Thanks for your edit and giving an example clearing the definition. – Ali Taghavi Aug 21 '19 at 20:52
• is it true (and obvious) that every every linear symplectomorphism of $\mathbb{R}^{2n}$ preserve $W$? – Ali Taghavi Aug 21 '19 at 20:54
• Is there an interesting NONLINEAR analogy for the spaces you are considering? – 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.
• 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 :) – darij grinberg Aug 22 '19 at 17:31
• @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. – Bugs Bunny Aug 22 '19 at 18:12