I heard someone say that representations of $\mathbb{Z}/2$ "are" super vector spaces. As far as I understand, super vector spaces are $\mathbb{Z}/2$-graded vector spaces, so my question is whether there is a canonical way of obtaining a $\mathbb{Z}/2$-graded vector space from a representation of $\mathbb{Z}/2$ (and vice versa).
Representations of $\mathbb{Z}/2$ as super vector spaces?
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linear-algebra
representation-theory
superalgebra
1 Answers
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Here's my best guess: let's write $\Bbb Z_2 = \{0,1\}$. A representation of $\Bbb Z_2$ consists of any assignment $\rho(1) = A:V \to V$ for which $A^2 = I$. Given such a linear transformation $A$, we have $$ V = \ker (A - I) \oplus \ker(A + I) $$ We would then take $V_0 = \ker(A - I)$ and $V_1 = \ker(A + I)$, in the notation of the wiki page.
Conversely, for any decomposition $V = V_0 \oplus V_1$, we could define a map $A$ by extending the definition $$ Av = \begin{cases} v & v \in V_0\\ -v & v \in V_1 \end{cases} $$
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0In a representation of a group, every element acts as an automorphism, so its kernel is zero... – 2017-01-28
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0Sorry, had projections on the brain – 2017-01-28
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0This seems to make sense, thanks. One can by the way describe this grading (at least for fields of characteristic $\neq 2$) also by decomposing the representation as $V \cong k_t^n \oplus k_s^m$, where $k_t$ is the trivial 1-dimensional representation and $k_s$ is the 1-dimensional sign representation, and declaring the "$k_t$ part" to be the even part and the "$k_s$ part" to be the odd part. – 2017-01-29