Let $k$ be a field, $L$ be a finite dimensional nilpotent Lie algebra over $k$ and $M$ be a finite dimensional irreducible representation of $L$. Assume that there is a linear function $\rho : L\rightarrow k$ such that $\forall h\in L, h-\rho(h)$ is a nilpotent linear transformation of $M$.

If the characteristic of $k$ is 0, one can verify that $M$ is 1-dimensional. This raises a natural question: If the characteristic of $k$ is $p$, is $M$ 1-dimensional?


1 Answer 1


This is VERY far from being true: consider the $3$-dimensional Heisenberg Lie algebra $L$ with basis $a$, $b$, $c$ and the only nonzero bracket $[a,b]=c$ (so $c$ is central in $L$). Consider the linear function $\rho$ on $L$ such that $\rho(c)=1$ and $a,b\in \ker\rho$. Assume the base field $K$ has characteristic $p>2$ and consider the $p$-dimensional vector space $V$ over $K$ with basis $v_0,v_1,\ldots, v_{p-1}$. Set $v_{-1}=v_p=0$ and define a linear action of $L$ on $V$ by setting $a.v_i=iv_{i-1}$, $b.v_i=v_{i+1}$ and $c.v_i=v_i$ for all $i$. It is straightforward to check that this is a $p$-dimensional irreducible representation of $L$.

Identify $a,b,c$ with their images in $\mathfrak{gl}(V)$. Then $a^p=b^p=0$ an $c^p=c={\rm Id}_V$. Since $p>2$, applying Jacobson't formula for $p$-th powers we get $$(\lambda a+ \mu b+\nu c-\nu {\rm Id}_V)^p=\lambda^p a^p+\mu^p b+\nu^pc-\nu^p{\rm Id}_V=\nu^pc-\nu^p {\rm Id}_V=0$$ for all $\lambda,\mu,\nu\in K$. It remains to note that $\rho(\lambda a+\mu b+\nu c)=\nu$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.