6
$\begingroup$

I cannot find any categorical definition of an eigenvalue, so I ask this question. Let $\mathbb{k}$ a be a field and $\mathcal{C}$ be a $\mathbb{k}$-linear abelian category. Let $f: X \rightarrow X \in \mathrm{End}_\mathcal{C}(X)$. To me, it makes sense to call $\lambda \in \mathbb{k}$ an eigenvalue of $f$ if $\ker(f - \lambda 1_X)$ is nonzero (and call this the corresponding eigenspace). By considering pullbacks one can show that these kernels do not "intersect" either for different $\lambda$.

If this indeed is the accepted definition, what are some reasonable set of conditions so that any such $f$ always has an eigenvalue (for instance, algebraic closedness of $\mathbb{k}$ will probably be necessary and some finiteness assumption)?

The greater context for such a question is from trying to prove categorical Schur's lemma for a tensor category, where any endomorphism of a simple object is a scalar multiple of the identity. And a similar statement about an endomorphism of an indecomposable being of the form $\lambda 1_X + n$, where $n$ is nilpotent.

$\endgroup$
7
$\begingroup$

Schur's lemma has the same proof in a $k$-linear abelian category $C$ as usual: if $T : M \to M$ is a nonzero endomorphism of a simple object, by simplicity it must have trivial kernel and cokernel, so is an isomorphism. Hence $\text{End}(M)$ is a division algebra over $k$. If furthermore $k$ is algebraically closed and $\text{End}(M)$ is finite-dimensional (e.g. if $C$ has finite-dimensional homsets) then $\text{End}(M) = k$.

Similarly if $k$ is algebraically closed and $\text{End}(M)$ is finite-dimensional then every endomorphism $T : M \to M$ has at least one eigenvalue (if $M$ is nonzero), because the natural map

$$k[x] \ni f(x) \mapsto f(T) \in \text{End}(M)$$

has nontrivial kernel (generated by the minimal polynomial of $T$). Working a little more carefully to check that all the details still work as usual without elements: if $m(t) = \prod (t - \lambda_i)^{m_i}$ is the minimal polynomial of $T$, then $m(T) = 0$ implies that (if $M \neq 0$) at least one of the factors $(T - \lambda_i)^{m_i}$ is not a monomorphism, hence has nontrivial kernel.

As for the indecomposable case, with the same hypotheses as above $M$ is naturally a module over $k[x]/m(x) \cong \prod k[x]/(x - \lambda_i)^{m_i}$. The primitive idempotents of this product split $M$ into the direct sum of generalized eigenspaces of $T$ (this is a general feature of idempotent endomorphisms in abelian categories and also does not require elements), so if $M$ is indecomposable then $T$ has exactly one eigenvalue $\lambda$ and $T - \lambda$ is nilpotent as usual.

| cite | improve this answer | |
$\endgroup$

Your Answer

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

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