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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

1 Answers

One Answer

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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