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Let $K$ be a field. Let $V$ be a finite-dimensional $K$-vector space; and let $f$ be an automorphism of $V$. Assume that $f$ is diagonalisable over some extension of $K$. Form the exterior algebra $\bigwedge^{*} V$. The automorphism $f$ induces an automorphism $\bigwedge^{*} f$ of $\bigwedge^{*} V$.

Now comes one technical assumption: assume that $f$ generates a (connected) torus in $\mathrm{GL}(V)$.

Can one reconstruct $f$ (up to conjugation) from $\bigwedge^{*} f$, viewing $\bigwedge^{*} V$ as $K$-vector space (i.e., forgetting the graded structure and the algebra structure)?

Observation: it suffices to prove this for algebraically closed fields.

In my specific case, I know slightly more, because I still know the $\mathbb{Z}/2\mathbb{Z}$-grading on $\bigwedge^{*} V$. In other words, I know $\bigwedge^{\text{even}} f$ and $\bigwedge^{\text{odd}} f$.

Motivation. I have a system of Galois representations $H_{\ell}$ , and I know that $\bigwedge^{\text{even}} H_{\ell}$ and $\bigwedge^{\text{odd}} H_{\ell}$ form compatible systems of weight $0$. Here compatibility means: the characteristic polynomials of Frobenii have coefficients in $\mathbb{Q}$, are independent of $\ell$, and their roots are $q$-Weil numbers. I would like to deduce that the $H_{\ell}$ form a compatible system themselves. Notably, my representations are motivic: they are subrepresentations of $l$-adic cohomology groups. So I know that eigenvalues of Frobenius are Weil numbers; but I do not know a priori that the characteristic polynomials have coefficients in $\mathbb{Q}$.

Rephrasing in primary school terms. Let $S$ be a finite multiset of complex numbers. (I assume your primary school treats basic set theory and complex numbers.) Define the $i$-th exterior multiset $$\bigwedge^{i} S = \big\{ \prod_{x \in X} x\, \big|\, X \subset S, \#X = i \big\}.$$ Put $\bigwedge^{\text{even}} S = \bigcup_{k \ge 0} \bigwedge^{2k} S$, and $\bigwedge^{\text{odd}} S = \bigcup_{k \ge 0} \bigwedge^{2k + 1} S$ (as multisets).

The technical condition stated above translates to: assume that all roots of unity in $S$ are equal to $1$.

Do the multisets $\bigwedge^{\text{even}} S$ and $\bigwedge^{\text{odd}} S$ determine the multiset $S$?

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  • $\begingroup$ What does "$f$ generates a torus" mean? Tori are algebraic subgroups; those usually aren't generated by a single element... $\endgroup$ Nov 7 '16 at 19:40
  • $\begingroup$ @user100824: If I understand your question correctly, the answer is no. Take $f \in SL_n$; then both $f$ and $f^{-1}$ generate the same ungraded action on $\wedge^* V$ (as can be seen by shifting the grading by $n$, then reversing the grading). Have I understood incorrectly? $\endgroup$
    – user44191
    Nov 7 '16 at 20:08
  • $\begingroup$ @darij grinberg: I'm guessing they mean that $f$ is a regular semisimple element of $GL_n$ which "generates" its centralizer. $\endgroup$
    – user44191
    Nov 7 '16 at 20:11
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The second claim is false, I believe. Take the multisets $\{1, -1, 1\}$ and $\{1, -1, -1\}$. Then we obtain $\Lambda^{odd}=\{1, -1, 1, -1\}=\{1, -1, -1, 1\}$ and $\Lambda^{even}=\{-1, -1, 1\}=\{-1, 1, -1\}$

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  • $\begingroup$ Thanks! That was embarrassingly easy. I'll have to think a bit harder then... $\endgroup$
    – user100824
    Nov 7 '16 at 19:08
  • $\begingroup$ Sorry for changing the rules in the middle of the game. But I realised that you provide a class of counterexamples that does not occur in my application... so I slightly changed the question. $\endgroup$
    – user100824
    Nov 7 '16 at 19:28
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If I've understood the question correctly, the answer should be no. A counterexample for the second question: $S = \{2, \frac{1}{3}, \frac{1}{4}, 6\}, S' = \{\frac{1}{2}, 3, 4, \frac{1}{6}\}$ both give $\Lambda^{odd} = S \cup S', \Lambda^{even} = \{1, 1, \frac{2}{3}, \frac{1}{2}, 12, \frac{1}{12}, \frac{3}{2}, 2\}$. More generally, let $S$ be a multiset such that $\prod S = 1, |S|$ even. Then $S, S^{-1} = \{s^{-1}|s \in S\}$ will have the same multisets. It's not hard to find examples that satisfy the technical condition and $S \neq S^{-1}$. This corresponds to the "inversion" I mentioned above.

More generally, if any even-sized subset of $S$ has product 1, then that subset can be "inverted", and the multisets will remain the same.

On the other hand: write $S = \{s_i\}$. Assume there are no $n_i \in \mathbb{Z}$ with $\prod s_i^{n_i} = 1$. Then you can determine $f$. There can even be bounds on the $n_i$ (I think it should be $|S|2^{|S|}$ or similar, but I'm not sure).

First, determine $\prod S$. This can be done, as $\prod \Lambda = (\prod S)^{2^{|S| - 1}}$, which determines $\prod S$ up to $\zeta_{2^{|S|} - 1}$, and if there were another element of $\Lambda$ with that power, then their quotient would be a root of unity - violating the condition. Then there is a unique set of $|S|$ elements of $\Lambda$ with product $|S|$, as if any other set had that product, finding a violation of the condition would be simple.

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