Is there a non-zero algebra homomorphism $\text{End}(V) \to \text{End}(\bigwedge^kV)$?

Question

This question is totally out of curiosity.

Let $V$ be a real $d$-dimensional vector space. Let $1<k<d$ be fixed.

Is there a non-zero algebra homomorphism $\text{End}(V) \to \text{End}(\bigwedge^kV)$?

Here $\bigwedge^kV$ is the $k$-th exterior power of $V$.

If there exist such a map $\phi$, then $\phi(\lambda \, id) \phi(A)=\phi(\lambda \, id \cdot A)=\phi(\lambda A)=\lambda \phi(A) $.

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Note that there are two familiar maps $\psi,\phi:\text{End}(V) \to \text{End}(\bigwedge^kV)$ respecting some of the structure:

1. Multiplicative: $\psi(A)=\bigwedge^kA$ (the exterior power functor).
2. Linear: $\phi=d\psi_{id}$ which is given by $A \to d\psi_{id}(A)$, where $$d\psi_{id}(A)(e_{i_1} \wedge \dots \wedge e_{i_k})= \sum_{s=1}^k e_{i_1} \wedge \dots \wedge Ae_{i_s}\wedge \dots \wedge e_{i_k}.$$

Answer 1

If $V$ is four-dimensional over a field $F$, then $\text{End}(V)\cong M_4(F)$ and $\text{End}(\bigwedge^2 V)\cong M_6(F)$. Is there an $F$-algebra homomorphism from $M_4(F)$ to $M_6(F)$?

I don't think so. The set of height-six column vectors is a left $M_6(F)$-module. An $F$-algebra homomorphism $\phi:M_4(F)\to M_6(F)$ would pull this back to a left $M_4(F)$-module, but up to isomorphism every simple $M_4(F)$-module is a four-dimensional $F$-vector space, so each $M_4(F)$ module's dimension as a vector space is a multiple of $4$.