# Differences in representations of Lie groups and Lie algebras

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.

• 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. Jan 6, 2022 at 1:39
• This is literally just the product rule.
– anon
Jan 6, 2022 at 1:58
• @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 Jan 6, 2022 at 7:38
• 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.]
– anon
Jan 6, 2022 at 16:39

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.