# Putting an $A otimes A^{mathrm{op}}$-module structure on $operatorname{Hom}(V,V')$

Observe that given an algebra $$A$$, we can put a $$A otimes A^{mathrm{op}}$$ structure on $$operatorname{Hom}(V,V’)$$ by $$((a otimes a’)f)(v)=af(a’v)$$

Indeed, we have
begin{align*} {}& ((a otimes a’)(b otimes b’))f(v) \ ={}& (ab otimes b’a’))f(v) \ ={}& ab(f(b’a’v)) \ ={}& a((b otimes b’)f(a’v)) \ ={}& (a otimes a’)((b otimes b’)f(v)) end{align*}

I’m just a bit confused why we need the $$A^{mathrm{op}}$$ rather than just $$A$$… Can we put an $$A otimes A$$ structure on $$operatorname{Hom}(V,V’)$$ by the same formula?

begin{align*} {}& ((a otimes a’)(b otimes b’))f(v) \ ={}& (ab otimes a’b’))f(v) \ ={}& ab(f(a’b’v)) \ ={}& (a otimes a’)bf(b’v)) \ ={}& (a otimes a’)((b otimes b’)f(v)) end{align*}

Maybe they both work… Thanks in advance!!

Mathematics Asked by A Dragon on December 28, 2020

Letting $$(af)(v) = f(av)$$ does not define an $$A$$-action, because it breaks associativity.

$$((ab)f)(v)$$ should equal $$f(abv)$$, but

$$(a(bf))(v) = (bf)(av) = f(bav)$$, reversing the multiplication of $$a$$ and $$b$$. This is why this is an $$A^{op}$$-structure.

Correct answer by Jacob FG on December 28, 2020

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