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Let $G$ be a lie group with lie algebra $\mathfrak{g}$ and let $M$ be a $G$-manifold. We denote the space of smooth differential forms on $M$ by $A(M)$. Let $E^i$ be a basis of $\mathfrak{g}$ and let $E_i$ $\in \mathfrak{g}^*$ be the dual basis. Let $S(\mathfrak{g}^*)$ be the symmetric algebra of $\mathfrak{g}^*$. Define the operator $\overline{d}$ on $S(\mathfrak{g}^*) \otimes A(M)$ by

$$\overline{d}(P \otimes \alpha)= \sum_i E_iP \otimes \iota (E^i_M) \alpha$$

for $P \in S(\mathfrak{g}^*)$ and $\alpha \in A(M)$ (note that $\iota (E^i_M)$ is the contraction by the vector field $E^i_M$ wich is defined by $$(E^i_M.f)(m)= \frac{d}{dt} \bigg|_{t=0} f(\exp(tE^i)m),$$ for $f \in C^\infty(M)$ and $m \in M$).

My question is why the definition of $\overline{d}$ is independent of the choice of the basis $E^i$ ?

Actually, I get this question by reading section $2$ of the article of equivariant cohomology with generalized coefficients by Kumar and Vergne, and the authors have commented on this by saying the following

This expression is independent of the choice of the basis $E^i$, as the element $\sum_i E_i \otimes E^i \in \mathfrak{g}^* \otimes \mathfrak{g}$ is the canonical element $I \in \operatorname{End}(\mathfrak{g})$, where $I$ is the identity element of $\operatorname{End}(\mathfrak{g})$.

But I didn't understand their comment! Moreover, I tried to prove this independence in the case where $\mathfrak{g}$ is one dimensional without going any where.

Please help with this.

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When $V$ is a finite-dimensional vector space over a field $k$, there is a canonical isomorphism of vector spaces $\varphi : V \otimes V^* \to \operatorname{End}_k(V)$, such that $v \otimes f$ gets taken to the linear operator $\varphi(v \otimes f): u \mapsto v f(u)$ and extended linearly. This map is not hard to understand in terms of a basis: if $e_1, \ldots, e_n$ is a basis of $V$ and $e_1^*, \ldots, e_n^*$ the dual basis, and we identify $\operatorname{End}_k(V)$ with $n \times n$ matrices using this basis, then $\varphi(e_i \otimes e_j^*)$ is the matrix with a $1$ in the $(i, j)$ place and zeros elsewhere.

Inside the vector space $\operatorname{End}_k(V)$ we have the identity map $\operatorname{id}_V \colon V \to V$, and we can ask what it corresponds to in $V \otimes V^*$. Many authors call this the canonical element $\varphi^{-1}(\operatorname{id}_V) \in V \otimes V^*$. Once we fix a basis and its dual, it is easy to check that $\varphi^{-1}(\operatorname{id}_V) = \sum_i e_i \otimes e_i^*$. Moreover, since $\varphi^{-1}(\operatorname{id}_V)$ was defined independent of any basis, it must be true for every basis and its dual that $\varphi^{-1}(\operatorname{id}_V) = \sum_i e_i \otimes e_i^*$.

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    $\begingroup$ thank you so much@Joppy for your helpful answer. $\endgroup$
    – asma
    Feb 5 at 11:11
  • $\begingroup$ Hi! I understand the answer, but I don't know how to use it to answer the question : why is the operator d independent of the choice of the basis? Could any one explain this please ? $\endgroup$
    – Maria
    Sep 3 at 5:47

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