I have two questions regarding the exterior derivative of vector valued forms when representations are involved:

Suppose $V$ is a vector space, $M$ a smooth manifold and $\omega$ is a $V$ valued $k$-form on $M$. Ie, $\omega \in \Omega^k(M;V)$. Suppose furthermore that $\rho_1:G\rightarrow GL(V)$ is a representation for some Lie Group $G$ and $\rho_2:\mathfrak{g}\rightarrow GL(V)$ is the induced Lie algebra representation.

The function $\rho_1(g)\circ \omega$ could be considered as a $V$ valued $k$-form on $M$. If $d:\Omega^k(M;V)\rightarrow \Omega^{k+1}(M;V)$ is the exterior derivative for $V$ valued $k$-forms, then

$d(\rho(g)\circ \omega)=\rho(g)\circ d\omega$.

I am just wondering why this is true?

Furthermore, in some lecture notes I'm reading the author also writes for $\eta\in \Omega^1(M;\mathfrak{g})$ that

$d(\rho_2(\eta)\circ \omega)=\rho_2(d\eta)\circ \omega-\rho_2(\eta)\wedge d\omega$

I am completely in the dark as to how this equation makes sense. I don't understand how the RHS is a 2-form or where the RHS even comes from. How would you make sense of this equation?


The first equation is true because $\rho(g)$ is a constant, so it commutes with the exterior derivative. More explicitly, let $\{e_i\}$ be a basis for $V$. Then you can write $\omega = \sum_i e_i \otimes \omega_i$ with $\omega_i \in \Omega^k(M)$. Also, write $\rho(g) e_i = \sum_j g_i^j e_j$. Then $$ d(\rho(g)\omega) = d(\sum_{ij} g_i^j e_j \otimes \omega_i) = \sum_{ij} g_i^j e_j \otimes d\omega_i = \rho(g) d\omega. $$

For the second equation you can do a similar thing (note that now $\eta$ is not constant and so you have to use the Leibniz rule). Note that $\rho_2(\eta)\circ \omega$ combines the wedge product on forms with the action of $\mathfrak g$ on $V$ and so is in $\Omega^{k+1}(M;V)$. So the right hand side should be a k+1 form, not a 2-form.


Some more details for the second: write $\eta = \sum_i X_i \otimes \eta_i$ where $X_i \in \mathfrak g, \eta_i \in \Omega^1(M)$. Write $\omega$ as before. By definition, $\rho_2(\eta) \circ \omega = \sum_{ij} (\rho_2(X_i)(e_j)) \otimes (\eta_i \wedge \omega_j)$. So $$ d(\rho_2(\eta) \circ \omega) = \sum_{ij} (\rho_2(X_i)(e_j)) \otimes d(\eta_i \wedge \omega_j) \\ = \sum_{ij} (\rho_2(X_i)(e_j)) \otimes (d\eta_i \wedge \omega_j - \eta_i \wedge d\omega_j) \\ = \sum_{ij} (\rho_2(X_i)(e_j)) \otimes d\eta_i \wedge \omega_j - \sum_{ij} (\rho_2(X_i)(e_j)) \otimes \eta_i \wedge d\omega_j \\ = \rho_2(d\eta) \circ \omega - \rho_2(\eta)\circ d\omega. $$ Hope this clears it up.

  • $\begingroup$ Thank you for your help with this (and all of my questions!) I really appreciate it. I follow what you wrote about the first equation but am still stuck on how to extend this to the case of the second equation... Do we start with $\eta=\sum_i \theta_i\otimes\eta_i$ where $(\theta_i)$ is a basis for $\mathfrak{g}$ and $\eta_i\in \Omega^1(M)$? $\endgroup$ – beedge89 Mar 16 '15 at 6:32
  • $\begingroup$ Also, I don't know how to interpret $\rho(d\eta)$... $\endgroup$ – beedge89 Mar 16 '15 at 6:36
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    $\begingroup$ Let me know if my edit helps. $\endgroup$ – Eric O. Korman Mar 16 '15 at 6:40
  • $\begingroup$ Yes. Absolutely. Thank you so much. I did not even understand the definition of $\rho_2(\eta)\circ \omega$ but you have really cleared all of this up for me. Thank you! $\endgroup$ – beedge89 Mar 16 '15 at 6:45
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    $\begingroup$ No problem! Notation in this area can be very confusing. I think most authors would write $\rho_2(\eta) \wedge \omega$ instead of $\rho_2(\eta) \circ \omega$. $\endgroup$ – Eric O. Korman Mar 16 '15 at 6:48

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