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) $.

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 $$d\psi_{id}(A)(e_{i_1} \wedge \dots \wedge e_{i_k})= \sum_{s=1}^k e_{i_1} \wedge \dots \wedge Ae_{i_s}\wedge \dots \wedge e_{i_k}.$$

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 $4$.

  • $\begingroup$ Thanks, this looks nice. Can you please clarify some points for me? (my abstract algebra is a bit rusty): (1) Why any simple $M_4(F)$-module is a 4-dimensional $F$-vector space (after isomorphism)? (2) Why does this imply any $M_4(F)$-module has dimension multiple of $4$? Is it because each module is a direct sum of simple modules? (Is that true for modules over arbitrary rings?). $\endgroup$ – Asaf Shachar Sep 3 '18 at 4:34
  • $\begingroup$ (3) Just to be sure I see why $\text{End}(\bigwedge^2 V), M_6(F)$ are isomorphic as $F$-algebras: The isomorphism I had in mind was the following: Let $\{ e_1,e_2,e_3,e_4\}$ be a basis for $V$. For $A \in \text{End}(\bigwedge^2 V)$ write $A(e_i \wedge e_j)=a^{ij}_{kl}e_k \wedge e_l$, then the map $A \to a^{ij}_{kl}$ is an isomorphism of algebras. Was this the isomorphism you had in mind? Thanks. $\endgroup$ – Asaf Shachar Sep 3 '18 at 4:34
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    $\begingroup$ $M_n(F)$ is a semisimple algebra: all its modules are direct sums of simple modules. Up to isomorphism, it only has one simple module, which has dimension $n$ over $F$. The endomorphism ring of an $n$-dimensional vector space over $F$ (as an $F$-algebra) is isomorphic to $M_m(F)$. $\endgroup$ – Lord Shark the Unknown Sep 3 '18 at 4:39
  • $\begingroup$ Thanks. Why does the fact the base ring is a semisimple algebra implies all its modules are direct sums of simple modules? Is it elementary? (do you know a reference?). I also do not see why this fact implies that there is one simple module up to isomorphism. $\endgroup$ – Asaf Shachar Sep 3 '18 at 4:47
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    $\begingroup$ @AsafShachar I learnt all about simple and semisimple algebras from Paul Cohn's textbook on algebra. $\endgroup$ – Lord Shark the Unknown Sep 3 '18 at 4:49

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