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Let $V$ be a finite dimensional vector space over $\mathbb{R}$. What can we say about the dimension of $V$ if we know that there exists some linear map $\phi: V\to V$ such that $\phi^n=-I$, where $I$ is the identity and $n>1$. Shouldn't we be able to infer structural information about our vector space based on such a map? Would you need to use representation theory to understand such a thing?

Edit: I'm supposing $n$ is the minimal such integer satisfying the above.

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up vote 2 down vote accepted

For the sake of completeness, here's what representation theory has to say (although since the relevant algebra is commutative it really becomes commutative algebra). You want to study finite-dimensional representations of (finitely-generated modules over) the algebra $R = \mathbb{R}[x]/(x^n + 1)$. Now, $\mathbb{R}[x]$ is a principal ideal domain, so by the structure theorem any such representation decomposes into a finite direct sum $$\bigoplus (R/f_i(x))^{e_i}$$

where $f_i(x)$ is an irreducible factor of $x^n + 1$ over $\mathbb{R}$. (One can also deduce this using Jordan normal form, but this is a special case of the structure theorem anyway.) Now, the identity $$x^n + 1 = \frac{x^{2n} - 1}{x^n - 1}$$

shows that the roots of $x^n + 1$ are the $2n^{th}$ roots of unity which are not $n^{th}$ roots of unity; these are precisely the roots of unity of the form $e^{ \frac{\pi i k}{n} }$ with $k$ odd. If $n$ is even, these come in complex conjugate pairs, and so all the $f_i$ are quadratic; if $n$ is odd, $x^n + 1$ has a unique linear factor $x + 1$ and the remaining $f_i$ are all quadratic.

Hence if $n$ is odd the finite-dimensional representations can have any finite dimension, and if $n$ is even the finite-dimensional representations can have any even dimension.

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If you want $n$ to be minimal, you just need to require that at least one of the $f_i$ has a primitive $2n^{th}$ root of unity as one of its roots. – Qiaochu Yuan Feb 10 '12 at 5:14
I was led to believe that we could obtain more, that for example if $n=2k$, we could say that $V=V_1\oplus\cdots\oplus V_{2k}$, where each $V_i$ is invariant under $\phi$. I may have misunderstood, or maybe representation theory isn't what I'm looking for. – user21725 Feb 10 '12 at 5:15
@Eric: well, each $R/f_i(x)$ is invariant under $\phi$ (that's part of what it means to be a direct summand as an $R$-module). But I don't know why you expect there to be $2k$ summands. There may be any positive integer number of irreducible summands. Perhaps by $V_i$ you mean the $i^{th}$ isotypic component? That is exactly the summand $(R/f_i(x))^{e_i}$ I used above. I'm not sure what more you are expecting exactly. – Qiaochu Yuan Feb 10 '12 at 5:23

If $n$ is odd, we can't conclude anything because then $\phi$ might be $-I$ itself. Or, more ambitiously, we can make $n$ be the least power such that $\phi^n=-I$ for any dimension $\ge 2$ by setting $$\phi=\begin{pmatrix}\cos(\pi/n)&\sin(\pi/n)\\-\sin(\pi/n)&\cos(\pi/n)\\ &&-1\\&&&\ddots\\&&&&-1\end{pmatrix}$$

On the other hand, if $n$ is even, then $(-1)^{\dim V}=\det(-I)=(\det \phi)^n$ which is positive, and therefore the dimension of $V$ is even. But in this case $V$ can still have any even dimension, by letting $\phi$ be a block diagonal matrix with $\begin{pmatrix}\cos(\pi/n)&\sin(\pi/n)\\-\sin(\pi/n)&\cos(\pi/n)\end{pmatrix}$ blocks on the diagonal.

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The question might be more interesting if one knew of an $n$ and a $\phi$ such that $\phi^n=-I$ but if $0\lt m\lt n$ then $\phi^n\ne-I$. – Gerry Myerson Feb 10 '12 at 4:47
Even then, the $\mathbb C^k$ example still works for even $n$, and for odd $n$ we could still have any dimension $k \ge 2$ (namely $V=\mathbb C\oplus \mathbb R^{k-2}$ and $\phi=e^{\pi i/n}\oplus -I_{k-2}$). – Henning Makholm Feb 10 '12 at 4:55
I should have written that $n$ was the minimal such integer. – user21725 Feb 10 '12 at 5:03
Taking the case where $n$ is even then, can't we say anything else about its structure? It seems the condition on $V$ is strong. – user21725 Feb 10 '12 at 5:04
Indeed. Henning, I ignored all mention of $\bf R$ and thought we were working over $\bf Q$. My apologies to all. – Gerry Myerson Feb 10 '12 at 5:06

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