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