Victor Kac, in the paper

"Classification of infinite-dimensional simple linearly compact Lie superalgebras", http://www.mat.univie.ac.at/~esiprpr/esi605.pdf

writes at the beginning of section 5 (p.39 in the linked preprint):

A consistent $\mathbb Z$-graded Lie superalgebra of depth $\geq 2$ is infinite-dimensional (since otherwise all even elements are exponentiable).

here a consistent $\mathbb Z$-graded Lie superalgebra of depth $d$ is a complex, possibly infinite dimensional, Lie superalgebra $$ \mathfrak g=\bigoplus_{p\in\mathbb Z,\ p\geq -d} \mathfrak g_p $$ with every $\mathfrak g_p$ finite dimensional, $\mathfrak{g}_{-d}\neq 0$, and $$ \mathfrak g_{\bar 0}=\bigoplus_{p\in\mathbb Z} \mathfrak g_{2p},\quad\mathfrak g_{\bar 1}=\bigoplus_{p\in\mathbb Z} \mathfrak g_{2p+1}. $$

The superalgebra $\mathfrak g$ is also supposed to be fundamental, i.e. $\mathfrak g_{-1}$ generates $\oplus_{p<0}\mathfrak g_p$, and transitive, i.e. $[X,\mathfrak g_{-1}]=0$ implies $X=0$ for $X\in\mathfrak g_p$, $p\geq 0$.

The statement is clearly true if $\mathfrak g$ is also even, i.e. if all even elements that are exponentiable have nonnegative degree. The result indeed is needed only for even superalgebras, however the hypothesis is not explicitly stated. My question is the following:

Are there finite-dimensional consistently graded fundamental transitive Lie superalgebras of depth $\geq 2$?



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy