What does $\lambda^2(\mathfrak{g}/\mathfrak{h})$ mean?

Question

I am reading Differential Geometry by Sharpe. Exercise 3.4.8 a) is:

Show that $\mathrm{Hom}(\lambda^2(\mathfrak{g}/\mathfrak{h}),\mathfrak{g})$ is an $H$ module under the action $$ (h\varphi)(v, w)=\mathrm{Ad}(h)\varphi(\mathrm{Ad}(h^{-1})v, \mathrm{Ad}(h^{-1})w),\quad h\in H. $$

(Here $H$ is a Lie group with Lie algebra $\mathfrak{h}$). What does the notation $\lambda^2(\mathfrak{g}/\mathfrak{h})$ mean?

It's definitely typed as a lowercase greek lambda. In several other places in the text as well. What would the wedge notation mean? — Andrea Allais, Jul 30 '20 at 15:27
Of course! That's right, I checked in other places, that's his notation for exterior power. @angina-seng If you want to add it as an answer I can accept it. Thanks! — Andrea Allais, Jul 30 '20 at 15:37

score 3 · Accepted Answer · answered Jul 30 '20 at 20:24

It seems to me that Sharpe uses both $\Lambda$ and $\lambda$ to denote exterior powers. For example, in 1.3.8 he uses the notation $\Lambda^p(V)$ and explicitely calls it the $p$-th exterior power. But in 1.4.17 he uses the notation $\lambda^n( T(M) )$ and explicitely calls this the $n$-th exterior power.

A popular notation for the $p$-th exterior power is $\bigwedge^p$, also already pointed out in the comments. Some people use $\Lambda^p$ instead of $\bigwedge^p$, so it’s understandable for Sharpe to use $\Lambda^p$. But I have no idea why he is also using $\lambda^p$. (Maybe this inconsistency should have been changed in editing the book but was overlooked?)

The assumption that $\lambda$ denotes an exterior power also seems to be supported by part (c) of the exercise. There Sharpe uses the notation $e_i^* \wedge e_j^* \otimes e_{kl}$ for an element of $\operatorname{Hom}(\lambda^2(\mathfrak{g}/\mathfrak{h}), \mathfrak{h})$. The simple wedges of an exterior power $\bigwedge^p(V)$ are commonly notated by $v_1 \wedge \dotsb \wedge v_n$, so the occurance of $\wedge$ also seems to suggest that $\lambda^2$ should be an exterior power.

Lastly, if $G$ is a group and $V$ and $W$ are two $G$-modules then $\operatorname{Hom}(\bigwedge^2(V), W)$ also becomes a $G$-module via $$ (g \varphi)(v_1 \wedge v_2) = g \varphi( (g^{-1} v_1) \wedge (g^{-1} v_2) ) \,. $$ If one identifies $\operatorname{Hom}(\bigwedge^2(V), W)$ with the space of alternating, bilinear maps from $V \times V$ to $W$ (via the universal property of the exterior power) then this action becomes $$ (g \varphi)(v_1, v_2) = g \varphi( g^{-1} v_1, g^{-1} v_2 ) \,. $$ And this looks a lot like the action that Sharpe proposes in the exercise.

What does $\lambda^2(\mathfrak{g}/\mathfrak{h})$ mean?

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