In the book Shun-Jen Cheng, Weiqiang Wang Dualities and Representations of Lie Superalgebrasm. One founds the following definition(Definition 1.3):

Let $\mathfrak{g}$ and $\mathfrak{g'}$ be Lie superalgebras. A homomorphism of Lie superalgebras is an even linear map $f: \mathfrak{g} \rightarrow \mathfrak{g'}$ satisfying $$f([a,b])=[f(a),f(b)],~ a, b \in \mathfrak{g}. ~~~~(*)$$

Here is my question:

Must a homomorphism of Lie superalgebras be even?

Assume $\mathfrak{g}$ is a Lie superalgebra, $A$ is a trivial $\mathfrak{g}$-supermodule. Then $A$ can be viewed as a Lie superalgebra with the zero superbracket. Let $\mathfrak{g'}=\mathfrak{g}\oplus A$ with $\mathfrak{g'}_{\bar 0}=\mathfrak{g}_{\bar 0}+A_{\bar 1}$, $\mathfrak{g'}_{\bar 1}=\mathfrak{g}_{\bar 1}+A_{\bar 0}$. Then $\mathfrak{g'}$ is a Lie superalgebra. We can define embeeding map $i: A \rightarrow \mathfrak{g'}=\mathfrak{g}\oplus A$, with the degree of $i$ is ${\bar 1}$. The map $i$ satisfies $$[i(a),i(b)]=i([a,b]).$$

Can we say this $i$ is a homomorphism of Lie superalgebras?

  • 2
    $\begingroup$ Most categories of superthings should admit a notion of "odd morphism." In the simplest case of supervector spaces this is because supervector spaces can be enriched over themselves. This is analogous to how chain complexes can be enriched over themselves, and consequently there is a natural notion of morphism of degree $i$, not necessarily zero, between chain complexes. $\endgroup$ Apr 2, 2014 at 5:59

1 Answer 1


It is possible to talk about odd homomorphisms as you have defined them, though an odd homomorphism will necessarily vanish on the subspaces $[\mathfrak{g}_{\overline{0}},\mathfrak{g}_{\overline{0}}]$ and $[\mathfrak{g}_{\overline{1}},\mathfrak{g}_{\overline{1}}]$ (assuming you are working over a field of characteristic not equal to $2$). In many interesting cases, this guarantees that the homomorphism will vanish on all of $\mathfrak{g}_{\overline{0}}$.

To see why the homomorphism will vanish on $[\mathfrak{g}_{\overline{0}},\mathfrak{g}_{\overline{0}}]$ and $[\mathfrak{g}_{\overline{1}},\mathfrak{g}_{\overline{1}}]$, let $\mathfrak{g}$ and $\mathfrak{g'}$ be Lie superalgebras, let $x$ and $y$ be homogeneous elements in $\mathfrak{g}$, and let $i: \mathfrak{g} \rightarrow \mathfrak{g}'$ be an odd homomorphism in the sense you have defined. Denote the homogeneous degrees of $x$ and $y$ by $d(x)$ and $d(y)$. Then $[x,y] = -(-1)^{d(x) \cdot d(y)} [y,x]$, so $$ i([x,y]) = -(-1)^{d(x) \cdot d(y)} i([y,x]). $$ Now using the fact that $i$ is a homomorphism, $$ i([y,x]) = [i(y),i(x)] = -(-1)^{d(i(x)) \cdot d(i(y))}[i(x),i(y)] = -(-1)^{d(i(x)) \cdot d(i(y))}i([x,y]). $$ Finally, $$(-1)^{d(i(x)) \cdot d(i(y))} = (-1)^{(d(x)+1)(d(y)+1)} = (-1)^{d(x)\cdot d(y)+d(x)+d(y)+1}$$ so it follows that $$ i([x,y]) = (-1)^{d(x)+d(y)+1}i([x,y]). $$ If $x$ and $y$ are both of the same parity, this implies that $i([x,y]) = -i([x,y])$, and hence (by the assumption that $1 \neq -1$ in the given field) that $i([x,y]) =0$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .