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Let $L$ be a (finite dimensional) Lie-algebra. Let $V, W$ be finite-dimensional vector spaces. If $V,\; W$ are in addition $L$-modules (see, e.g., 6.1 in Humphreys Introduction to Lie Algebras), then using the isomorphism $V^{*}\otimes W\simeq Hom\left(V,W\right)$, we can define an $L$-module structure on $Hom\left(V,W\right)$ by $x.\left(f\otimes w\right)\left(v\right):=\left(f\otimes x.w+x.f\otimes w\right)\left(v\right)=f\left(v\right)\otimes x.w-f\left(x.v\right)\otimes w$ for $x\in L$, $v\in V$, $f\in V^{*}$. This can be written explicitly as

$x.T\left(v\right)=\sum_{i}x.\left(v_{i}^{*}\otimes Tv_{i}\right)\left(v\right)=\sum_{i}\left[v_{i}^{*}\left(v\right)\; x.Tv_{i}-v_{i}^{*}\left(x.v\right)\; Tv_{i}\right]$

for $T\in Hom\left(V,W\right)$, $\left(v_{i}\right)_{i}$ a basis for V.

Question: The above equation has a clear asymmetry (minus sign) between the action of $L$ on the domain (-) and its action on the codomain (+). Having such an asymmetry is surprising. Technically, it comes from the definition of the $L$-modules structure $\left(x.f\right)\left(v\right):=-f\left(x.v\right)$ for the dual of the $L$-module $V$. Yet, this is merely a technical explanation for the above asymmetry. Why do we have this asymmetry?

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    $\begingroup$ Because you want the action of the associated Lie group to be the obvious one, treating $V \otimes W^*$ as having dual action of $W$ on the $W^*$ factor. $\endgroup$
    – Ben McKay
    Commented Apr 11, 2016 at 20:21
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    $\begingroup$ Why would you expect symmetry? You are describing the action on $V^* \otimes W$ in terms of the actions on $V$ and $W$. $\endgroup$ Commented Apr 11, 2016 at 22:33
  • $\begingroup$ @josé-figueroa-ofarrill, that's a good point. Perhaps symmetry shouldn't be expected here. Thanks! $\endgroup$
    – Shlomi A
    Commented Apr 12, 2016 at 4:50

1 Answer 1

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If $V$ is a representation of a group $G$ then the dual space $V^*$ is naturally a $G^{\text{op}}$-representation. To get a $G$-action use inversion. For a Lie algebra, the analogue of inversion is $x \mapsto -x$.

To see this fact about groups from a more general perspective: let $BG$ be the groupoid whose one object has automorphism group $G$. Then an action of $G$ on an object $X$ of a category $\mathscr{C}$ is a functor $BG \to \mathscr{C}$ which sends the unique object of $BG$ to $X$. If $F : \mathscr{C}^{\text{op}} \to \mathscr{D}$ is a contravariant functor then we obtain an action $$B(G^{\text{op}}) \to (BG)^{\text{op}} \to \mathscr{C}^{\text{op}} \to \mathscr{D}$$ of $G^{\text{op}}$ on $F(X)$. Now apply this with $\mathscr{C} = \mathscr{D}$ the category of vector spaces and $F$ the duality functor.

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    $\begingroup$ How much obscure is your second paragraph compared to the first one! $\endgroup$ Commented Apr 11, 2016 at 21:48
  • $\begingroup$ The point is that contravariance forces you to write an inverse if you want to keep a left action, and which when differentiated becomes a minus sign. Look, no formulas! $\endgroup$ Commented Apr 11, 2016 at 21:57
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    $\begingroup$ That sentence in the comment sheds more light on the matter than 1729 mentions of the word groupoid :-) $\endgroup$ Commented Apr 11, 2016 at 21:59
  • $\begingroup$ This is a great answer; I always enjoy categorical explanations $\endgroup$
    – ಠ_ಠ
    Commented Apr 11, 2016 at 22:41

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