Grammatically confused: $\omega=4dV$ for 3-form $\omega$ and volume in $\Bbb R^4$?

Question

Background: Against the advice I should have been given but wasn't, I'm taking a Lie theory course with no background in differential geometry. We finally made it into the part of the course where we need to know differential forms, and I'm not taking it very well.

I ran into a notational confusion on a homework problem. I'm also not really sure how to start it, but I'd like to think on that for myself a while.

Here is the full homework problem (it is exercise 4.10 from Kirillov's Introduction to Lie Groups and Lie Algebras.)

$$\DeclareMathOperator{\SU}{SU} \DeclareMathOperator{\su}{\mathfrak{su}} \DeclareMathOperator{\tr}{tr} \newcommand{\homeo}{\cong} \newcommand{\ss}{\subseteq} $$

Let $G = \SU(2)$. Recall that we have a diffeomorphism $G\homeo S^3$ (see Example 2.5).

• Let $\omega\in \Omega^3(G)$ be a left-invariant 3-form whose value at $1\in G$ is defined by $\omega(x_1, x_2, x_3) = \tr([x_1, x_2]x_3), x_i\in\mathfrak{g}$. [...] Show that $\omega = \pm 4dV$ where $dV$ is the volume form on $S^3$ induced by the standard metric in $\Bbb R^4$ [...]

• Show that $\frac{1}{8\pi^2}\omega$ is a bi-invariant form on $G$ such that for appropriate choice of orientation on $G$, $$\frac{1}{8\pi^2}\int_G\omega=1.$$

The first question doesn't make sense to me; isn't $dV$ supposed to be a 4-form $dx^1\land dx^2\land dx^3\land dx^4$?

(This is probably a different question, but I'm having a hard time understanding what does the wedge product mean at the pointwise level? For instance, with the volume form: $x^i$ is the function that picks out a coordinate $x\mapsto x_i$. Okay, so take the derivative of that; this is supposed to basically be the vector $x_ie_i\in\Bbb R^4$ at all $p$, yes? Is it then reasonable to get $(dV)_p(x)$ by shoving vectors into a matrix and... taking the determinant? This "works", in that it gets the right answer and explains the antisymmetry, but I'm worried that this intuition breaks for more general top-dimensional differential forms.)

Answer 1

Hint: the form $dV$ is defined on the Lie algebra $su(2)$ of $SU(2)$ you can represent the elements of $su(2)$ by complex $2\times 2$-matrices and use the trace and Lie bracket.

$su(2)=\pmatrix{ia & -c+id\cr c+id & -ia}a,b,c\in R.$ thus $su(2)$

$SU(2)=\pmatrix{a & -\bar b\cr b & \bar a}, a,b\in C \mid a\mid^2+\mid b\mid^2=1$ thus it is a subspace of $R^4$ to evaluate $dV$ on $u,v,w$ tangent to $x\in SU(2)$, it is $dv(x,u,v,w)$

Answer 2

In $\Bbb{R}^4$, certainly, the volume form is $dV = dx^1 \wedge dx^2 \wedge dx^3 \wedge dx^4$. However, confined to the sphere $S^3$ in $\Bbb{R}^4$, you need a $3$-form. (It is perhaps easier to see this for the sphere $S^2$ in $\Bbb{R}^3$, where what is wanted is for areas, not volumes.)