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I am currently reading Mathematical Gauge Theory by Hamilton, and have come across something I cannot seem to reconcile. Say we have Lie group $G$ and a representation $\rho_{\Lambda^k V}$:

$$G\rightarrow \text{GL}\left(\Lambda^k V\right)$$

Then the book says that $\rho_{\Lambda^kV}(g)=\hat{g}$ acts on some $\omega\in\Lambda^k V$ by: $$\rho_{\Lambda^kV}(g)(\omega)=\hat{g}v^1\wedge\cdots \wedge gv^n$$

This makes a lot of sense, since if $\hat{g}$ can be written as a matrix then we have that previous line goes to:

$$\hat{g}(\omega)=\det{g}\cdot\omega$$

Which is what I expect based on my experience with tensor products and alternating forms. However, say $\phi_{\Lambda^k V}$ is a representation of Lie algebra $\mathfrak{g}$ of $G$ such that: $$\phi_{\Lambda^k V}: \mathfrak{g}\rightarrow \text{End}\left(\Lambda^k V\right)$$ then the book says that $\phi_{\Lambda^k V}(X)=\hat{X}$ acts on some $\omega \in \Lambda^k V$ by: $$X\omega=\sum_i v^1\wedge\cdots\wedge X v^i \wedge\cdots \wedge gv^n$$

Which I do not get at all. It seems to me that this might be a way to circumvent the situation where $Xv^i= 0$ so that our $\omega$ doesn't just get mapped to zero, but I can't see the justification here, or where this coming from at all. I suspect I missing something obviously subtle that I didn't pick up my on my first go around dealing with alternating forms.

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    $\begingroup$ I don't have time to write something out in depth, but you should examine how a lie group representation $G\to \mathrm{GL}(V)$ induces a lie algebra representation $\mathfrak{g}\to \mathfrak{gl}(V)$. This is done by taking a path $\gamma(t)$ through $e$ in $G$ with $\gamma'(0)=X\in \mathfrak{g}$ and defining $Xv=\lim_{t\to 0}\frac{\gamma(t)v-\gamma(0)v}{t}$, i.e. by differentiation. What you are seeing is then a version of the product rule. $\endgroup$ Jan 6, 2022 at 1:39
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    $\begingroup$ This is literally just the product rule. $\endgroup$
    – anon
    Jan 6, 2022 at 1:58
  • $\begingroup$ @runway44 I'm sorry I've only just started going in depth into Lie groups, how is this just product rule? Could you show me where to start to see that $\endgroup$
    – Chris
    Jan 6, 2022 at 7:38
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    $\begingroup$ Start with Alexos comment. If you differentiate $\gamma(t)v^1\wedge\cdots\gamma(t)v^n$ with respect to $t$ and evaluate at $t=0$, what do you get? To take the derivative, you use the product rule (which works since differentiation is linear and $\wedge$ is (bi/multi)linear). [The proof of the usual product rule works to prove the version of the product rule used here.] $\endgroup$
    – anon
    Jan 6, 2022 at 16:39

1 Answer 1

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Alright so the proof goes something like this I guess:

Let $\omega\in \Lambda^kV$ then we can write:

$$\omega=ae^1\wedge\cdots \wedge e^k$$

for some $a\in\mathbb{R}$, where $\{e^i\}$ form a basis of $V$. Let $\gamma(t)$ be a curve in a Lie group $G$ such that $\gamma(0)=0$, and $\dot{\gamma(0)}=X\in \mathfrak{g}$. Now let:

$$F:G\rightarrow \text{GL}(\Lambda^k)$$

be defined by: $$g\mapsto a\cdot ge^1\wedge\cdots\wedge ge^1$$

We then have that: $$dF_e(X)=a\cdot \frac{d}{dt}\Bigg|_{t=0} \gamma(t)e^1\wedge \cdots \wedge \gamma(t)e^k$$ $$=a\cdot\lim_{t\rightarrow 0}\sum_{i=1}^k \gamma(t)e^1\wedge\cdots \wedge \dot{\gamma}(t)e^i\wedge \cdots \wedge \gamma(t)e^k$$ $$=a\cdot\sum_{i=1}^k e^1\wedge\cdots \wedge X e^i\wedge \cdots \wedge e^k$$

As desired.

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