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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.

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  • $\begingroup$ 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. $\endgroup$
    – Kurt G.
    Commented Feb 7 at 11:48
  • $\begingroup$ I'm sorry but I don't see how you are constructing $dx$, $dy$, $dz$ the same way as in my post. $\endgroup$
    – user920957
    Commented Feb 7 at 12:07
  • $\begingroup$ Is $dx,dy,dz$ not a dual basis of $\mathbb R^3\,?$ Here is another elegant approach to $SU(2)\,.$ $\endgroup$
    – Kurt G.
    Commented Feb 7 at 14:27

1 Answer 1

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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 $\omega_1\wedge\dots\wedge\omega_n=vol_G$.

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  • $\begingroup$ Thank you. So if my group is non-compact and I just want a Haar measure (not necessarily normalized in any way) then my construction works. It's just my notation that is bad, which I see now. $\endgroup$
    – user920957
    Commented Feb 11 at 9:01
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    $\begingroup$ 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. $\endgroup$ Commented Feb 11 at 19:51

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