# Relating integration of forms to Haar measure integration on a Lie group

Let $$G$$ be a Lie group with Haar measure $$\mu$$ and Lie algebra $$\mathfrak{g}$$ given as the space of left-invariant vector fields on $$G$$. I want to understand the relationship between integration of forms on $$G$$ and integration with respect to Haar measure. Any reference on this would be highly appreciated.

Let me give a guess on such a relation. Choose a basis $$X_1, \ldots, X_n$$ of $$\mathfrak{g}$$. Let $$dx_1, \ldots, dx_n$$ be the dual basis in $$\Gamma(G, T^*G) = \Omega^1(G)$$, i.e. over every point of $$G$$ we have $$dx_i(X_j) = \delta_{ij}$$. My guess is then that $$dx_1 \wedge \cdots \wedge dx_n$$ is a non-vanishing form and that we should have something like

$$\displaystyle \int f \, dx_1 \wedge \cdots \wedge dx_n = \int f \, d\mu$$

for every smooth $$f \colon G \to \mathbb{R}$$. But since this depends on a choice of basis for $$\mathfrak{g}$$ I'm not sure equality is the right guess.

Edit: Having thought some more about this a possible way to prove or disprove my guess is to check if

$$\displaystyle \int f(yx) \, dx_1 \wedge \cdots \wedge dx_n = \int f (x) \, dx_1 \wedge \cdots \wedge dx_n$$

for all $$y \in G$$ and $$f$$ smooth and compactly supported. This seems plausible to me.

• Have you tried this with $SU(2)\,?$ A basis of ${\mathfrak su}(2)\cong\mathbb R^3$ is -as we know- $dx,dy,dz\,.$ If the Haar measure $\mu$ were $dx\wedge dy\wedge dz$ the volume of the $3$-sphere would be $\int_{S^3}dx\wedge dy\wedge dz$ which cannot be true. Commented Feb 7 at 11:48
• I'm sorry but I don't see how you are constructing $dx$, $dy$, $dz$ the same way as in my post. Commented Feb 7 at 12:07
• Is $dx,dy,dz$ not a dual basis of $\mathbb R^3\,?$ Here is another elegant approach to $SU(2)\,.$ Commented Feb 7 at 14:27

You can avoid the problem of choosing a basis by looking directly at the space of alternating $$n$$-linear maps $$\mathfrak g^n\to\mathbb R$$, where $$n=\dim(\mathfrak g)$$. By linear algebra, the space of all such maps is one-dimensional. Choosing an element in there defines a left invariant $$n$$-form on $$G$$ (by transporting tangent vectors to the neutral element $$e$$ by left translation and then plugging the resulting elements of $$\mathfrak g$$ into the map). For compact $$G$$, this form can be integrated over $$G$$ and it is easy to see that the integral is non-zero, so the initial multilinear map can be uniquely rescaled in such a way that the integral is one (which is the usual normalization for Haar measure). So you get an $$n$$-linear map $$\alpha:\mathfrak g^n\to\mathbb R$$ such that the corresponing invariant $$n$$-form $$vol_G$$ satisfies $$\int_Gvol_G=1$$.
You can easily connect this to what your were trying, you just have to require that the basis $$X_1,\dots,X_n$$ of $$\mathfrak g$$ that you choose satisfies $$\alpha(X_1,\dots,X_n)=1$$. Writing the dual basis as $$dx_1,\dots,dx_n$$ may be a bit misleading, since these are not coordinate $$1$$-forms in general (in particular, they are not closed in general), so I'll write them as $$\omega_1,\dots,\omega_n$$. Anyway, the definition implies that $$(\omega_1\wedge\dots\wedge\omega_n)(e)(X_1,\dots,X_n)=1=\alpha(X_1,\dots,X_n)$$ thus $$(\omega_1\wedge\dots\wedge\omega_n)(e)=\alpha$$ by dimension one. Since both $$a\omega_1\wedge\dots\wedge\omega_n$$ and $$vol_G$$ are left invariant, this implies $$\omega_1\wedge\dots\wedge\omega_n=vol_G$$.
• Yes, you can always use a basis of left invariant vector fields to construct a left invariant $n$-form (which then is unique up to a constant multiple). And you can use this to define a left invariant integral on compactly supported functions on your group. Commented Feb 11 at 19:51