# Non-graded representations over Lie superalgebra $\mathfrak{gl}(m,n)$

I have the following questions: Let $m,n$ be positive integers. Consider representations over the general linear Lie super-algebra $\mathfrak{gl}(m,n)$. Namely, modules over the associative algebra $U(\mathfrak{gl}(m,n))$.

Recall that one always considers the usual notion: "The graded representations" are those modules which have $\mathbb{Z}_2-$gradation compatible with the action of $\mathfrak{gl}(m,n)$. A representation is said to be non-graded if it is not a graded representation.

$\bf My$ $\bf Questions:$ Is there a finite-dimensional, non-graded representation over $\mathfrak{gl}(m,n)$? How can we construct such representation in general?

Thanks very much!

• It's not helpful to formulate more than one question, but if you need to do that it's best to number them. Also, you should be more precise about the use of "positive". – Jim Humphreys Jul 9 '15 at 15:43

The answer to your first question is "yes." Here is a particularly simple-minded example.

Suppose $n=0$, so that $\mathfrak{gl}(m,n) = \mathfrak{gl}(m)$. Now if $V$ is a graded representation of $\mathfrak{gl}(m)$, then it must be the case that $V = V_0 \oplus V_1$ for some $\mathfrak{gl}(m)$-submodules $V_0$ and $V_1$ of $V$, i.e., $V_0$ is the even subspace of $V$ and $V_1$ is the odd subspace of $V$, and these must both be $\mathfrak{gl}(m)$-submodules of $V$ because $\mathfrak{gl}(m)$ is a purely even Lie (super)algebra.

I will assume that we are working over a field $k$ of characteristic not dividing $m$. Then $\mathfrak{gl}(m)$ is isomorphic as a Lie algebra to $\mathfrak{sl}(m) \oplus k \cdot I_m$, where $k \cdot I_m$ denotes the one-dimensional abelian Lie subalgebra of $\mathfrak{gl}(m)$ spanned by the identity matrix (and of course $\mathfrak{sl}(m)$ denotes the subalgebra of trace-zero matrices in $\mathfrak{gl}(m)$).

Now let $V$ be a two-dimensional $k$-vector space with basis $\{v_0,v_1\}$. Consider the subspace of $V$ spanned by $v_0$ to be the even subspace of $V$, and the subspace of $V$ spanned by $v_1$ to be the odd subspace of $V$. We can make $V$ into a $\mathfrak{gl}(m)$-module by having $\mathfrak{sl}(m)$ act trivially on $V$, and by having $I_m.v_0 = v_1$ and $I_m . v_1 = 0$. Now $V$ is not a graded representation of $\mathfrak{gl}(m)$, because the vector space decomposition of $V$ into its even and odd subspaces is not a module direct sum decomposition.

Incidentally, this example is perhaps also the canonical example showing that not every finite-dimensional representation of a reductive Lie algebra need be semisimple.

EDIT: I overlooked the "positive" assumption in the question, so here is an example more of the type the OP was looking for.

Consider the Lie superalgebra $\mathfrak{gl}(1,1)$. I will assume that we are working over a field of characteristic not equal to $2$. Then $\mathfrak{gl}(1,1)$ admits a basis $\{e,f,I,h\}$ such that $e$ and $f$ are odd, $I$ and $h$ are even, and the only non-trivial commutator relations between these basis vector are $$[e,f]=I,$$ $$[h,e]=2e,$$ $$[h,f]=-2f.$$ Now let $V$ be a two-dimensional $\mathbb{Z}_2$-graded vector space as I defined in my previous example. We can make $V$ into a module for $\mathfrak{gl}(1,1)$ by having $e$, $f$, and $I$ act trivially on $V$, and by having the even element $h$ act by $h.v_0 = v_1$ and $h.v_1 = 0$. As above, this is not a graded representation for $\mathfrak{gl}(1,1)$, since the even subalgebra $\mathfrak{gl}(1,1)_0$ does not act by $\mathbb{Z}_2$-degree preserving endomorphisms.

• The assumption was that $m,n$ are both "positive" (which often means strictly positive), so your type of example may not suffice here. – Jim Humphreys Jul 9 '15 at 15:41
• I have updated my answer to reflect the "positive" assumption. – Christopher Drupieski Jul 9 '15 at 19:14