# Action of $L$ on $End(V)$.

I'm reading Introduction ot Lie Algebras and Representation Theory from James Humphreys and I do not understand the statement made at the top of page 27.

Given a vectorspace $V$ (finite dimensional) over a field $F$ (char 0), there is a standard isomorphism of vector spaces $V^* \otimes V \rightarrow End(V)$ degined by $(f \otimes v)(w) = f(w)v$. Now if $V$ is an $L$-module ($L$ a lie algebra) then $End(V)$ can be viewed as an $L$-module by $(x.f)(v) = x.f(v) - f(x.v)$.

I want to verify this using the isomorphism of $V^* \otimes V$ and $End(V)$. However, I got stuck in the calculations. This is what I did:

Let $f(v) = \sum \lambda_{ij} e_i^*(v) e_j$ be an endomorphism of $V$ and $x \in L$. Then $(x, f)$ in $L \times End(V)$ can be identified with $$(x, \sum \lambda_{ij} e_i^* \otimes e_j)$$ in $L \times (V^* \otimes V)$. "Differentiating" this element gives $$\sum \lambda_{ij} \left( (x.e_i^*) \otimes e_j + e_i^* \otimes (x.e_j) \right)$$ in $V^* \otimes V$. Then this isomorphism discribed above evaluated in a vector $v \in V$ gives \begin{align} \sum \lambda_{ij} (x.e_i^*(v)e_j + e_i^*(v)(x.e_j)) &= - \sum \lambda_{ij} e_i^*(x.v)e_j + \sum \lambda_{ij} e_i^*(v)(x.e_j) \\ &= \sum \lambda_{ij} e_i^*(v)(x.e_j) - f(x.v). \end{align}

I cannot manage to simplify this anymore. So I probably made a mistake somewhere. Could someone show me the correct calculations?

Edit: I think I found the answer already! Since $e_i^*(v) \in F$ it can be put trough the action $x$, hence $$\sum \lambda_{ij} e_i^*(v)(x.e_j) - f(x.v) = x.(\sum \lambda_{ij} e_i^*(v)e_j) - f(x.v) = x.f(v) - f(x.v).$$ I think this is correct?

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Consider the map $\psi:V\otimes V^* \to \mathrm{End}(V)$ defined by (for all $w \in V$) $\psi\left(\sum\limits_i v_i \otimes f_i\right)(w) = \sum\limits_i f_i(w)v_i$.

• You already know $\psi$ is a linear isomorphism

• Suppose $x\in L$ and let $w \in V$, then $\psi\left(x \cdot (v \otimes f) \right)(w)$ $= \psi\left((x \cdot v) \otimes f + v \otimes (x \cdot f) \right)(w)$ $= f(w)(x \cdot v)+(x \cdot f)(w)v$ $= f(w)(x \cdot v)-f(x\cdot w)v$.

On the other hand, $(x \cdot \psi(v \otimes f))(w) = (x \cdot (f(\cdot)v))(w) = x \cdot (f(w)v) - f(x \cdot w)v = f(w)(x \cdot v) - f(x \cdot w)v$.

Since these match for arbitrary $w \in V$, we have $\psi\left(x \cdot (v \otimes f) \right) = x \cdot \psi(v \otimes f)$. This holds on pure (or simple) tensor elements, and (since everything is linear) holds on arbitrary tensors.

Therefore, $\psi$ is an $L$-module isomorphism. Thus $V\otimes V^* \cong \mathrm{End}(V)$ as $L$-modules.

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Thank you for your answer, but I think I was asking something different. I'm trying to derive the action $(x.f)(v) = x.f(v) - f(x.v)$ from the map $L \times End(V) \rightarrow L \times (V^* \otimes V) \rightarrow (V^* \otimes V) \rightarrow End(V)$. However, I think that I already found out how to do this. – Noud Jan 12 '12 at 16:03