a) Denote by $u_1\in$ Alt$^s(M)$ the form $(x_1, . . . , x_s) \rightarrow u(p(x_1), . . . , p(x_s))$, and let $v_1\in$ Alt$^r(M)$ be a form such that $v_1(i(x'_1), . . . , i(x'_r)) =v(x'_1, . . . , x'_r)$ for $x'_1, . . . , x'_r$ in $M'$; then $u\cap v=u_1\wedge v_1$.

b) For all $x_1, . . . , x_s$ in M and $x'_1, . . . , x'_r$ in $M'$,

The map $\varphi :$ Alt$^s(M'')\otimes_A$Alt$^r(M')\rightarrow$ Alt$^{s+r}(M)$ such that $\varphi (u\otimes v) =u\cap v$ is an isomorphism of free A-modules of rank one.

The existence of a form $v_1$ satisfying condition a) follows from the fact that $\wedge \wedge^r(i)$ induces an isomorphism from $\wedge \wedge^r(M')$ to a direct factor submodule of $\wedge \wedge^r(M) ($Algebra, Chap. III, §7, no. 2). Let $v_1$ be such a form; put $u\cap v=u_1\wedge v_1$. Formula (1) is then satisfied, since if we put $i(x'_k) =x_{s+k}$ for $1\leq k\leq r$, the only element $\sigma$ of $\mathfrak{S}_{s,r}$ such that $p(x_{\sigma(i)})\not= 0$ for $1\leq i\leq s$ is the identity permutation. On the other hand, formula (1) determines $u\cap v$ uniquely: indeed, let $(e'_1, . . . , e'_r)$ be a basis of $M', (f''_1, . . . , f''_s)$ a basis of $M''$, and $f_1, . . . , f_s$ elements of M such that $p(f_i) =f''_i$ for $1\leq i\leq s$. Then $(f_1, . . . , f_s, i(e'_1), . . . , i(e'_r))$ is a basis of M (Algebra, Chap. II, §1, no. 11, Prop. 21), and formula (1) can be written

but an element of Alt$^{s+r}(M)$ is determined by its value on a basis.

It follows from the preceding that each of the conditions a) and b) determines the product $u\cap v$ uniquely; it is clear that this product is bilinear. Finally, the last assertion of the lemma follows from formula (2).

Clearly we can assume that $\omega_{G/T}$ and $\omega_T$ are non-zero; then the differential form $(u, t) \rightarrow \omega_{G/T}(u)\wedge \omega_T(t)$ on $(G/T)\times T$ is of degree $n$ and everywhere non-zero; hence there exists a numerical function $\delta$ on $(G/T)\times T$ such that

Observe now that, for $h\in G,u\in G/T,t\in$ T, we have $f(h.u, t) =$ (Int $h$)$f(u, t)$; since $\omega_G$ is invariant under inner automorphisms, it follows immediately that $\delta (h.u, t) =\delta (u, t)$, so $\delta (u, t) =\delta (\overline{e}, t)$.

Denote by $p:\mathfrak{g}\rightarrow \mathfrak{g}/\mathfrak{t}$ the quotient map and by $\varphi :\mathfrak{g}/\mathfrak{t}\rightarrow \mathfrak{g}$ the map defined by

$\varphi (p(X)) =$ (Ad $t^{-1}$)$X-X$ for $X\in \mathfrak{g}$; recall (§5, no. 4, Lemma 4) that the tangent map

takes $(z, tH)$ to $t(\varphi (z) +H)$ for $z\in \mathfrak{g}/\mathfrak{t}, H\in \mathfrak{t}$.

Let $z_1, . . . , z_{n-r}$ be elements of $\mathfrak{g}/\mathfrak{t},H_1, . . . , H_r$ elements of $\mathfrak{t}$. Then

$=\omega_G(t)(t\varphi (z_1), . . . , t\varphi (z_{n-r}), tH_1, . . . , tH_r)$ by the calculation of $T_{(\overline{e},t)}(f)$

$=\omega_G(e)(\varphi (z_1), . . . , \varphi (z_{n-r}), H_1, . . . , H_r)$ since $\omega_G$ is invariant

$=\omega_{G/T}(\overline{e})(p\varphi (z_1), . . . , p\varphi (z_{n-r})).\omega_T(e)(H_1, . . . , H_r)$ (no$.1$, Lemma 1)

= det($p\varphi$ )$\omega_{G/T}(\overline{e})(z_1, . . . , z_{n-r}).\omega_T(e)(H_1, . . . , H_r)$

since $\omega_T$ is invariant

so $f^*\omega_G(\overline{e}, t) =\delta_G(t)(\omega_{G/T}\wedge \omega_T)(\overline{e}, t)$; thus, $\delta (\overline{e}, t) =\delta_G(t)$, hence the proposition.

Give the manifolds $G,T$ and $G/T$ the orientations defined by the forms $\omega_G, \omega_T$ and $\omega_{G/T}$, respectively. These forms define invariant measures on $G,T$ and $G/T$ (Chap. III, §3, no. 16, Props. 55 and 56), also denoted by $\omega_G, \omega_T$ and $\omega_{G/T}$.

Denote by $\pi$ the canonical morphism from G to $G/T$. Let $g\in G$, and let $t_1, . . . , t_{n-r}$ be elements of $T_{\pi(g)}(G/T)$. Identify the fibre $\pi^{-1}(\pi (g)) =gT$ with T by the translation $\gamma (g)$. The relation $\omega_G=\omega_{G/T}\cap \omega_T$ now implies the equality (Differentiable and Analytic Manifolds, Results, 11.4.5):

Thus $\int_{\pi}\omega_G=(\int_T\omega_T)\omega_{G/T}($Differentiable and Analytic Manifolds, Results, 11.4.6), and

(Differentiable and Analytic Manifolds, Results, 11.4.8).

Choose an invariant differential form $\omega_T$ (resp. $\omega_{G/T}$) on T (resp. $G/T$) of maximal degree, such that the measure defined by $\omega_T$ (resp. $\omega_{G/T}$) is equal to $dt$ (resp. $\mu$). Put $\omega_G=\omega_{G/T}\cap \omega_T$. Lemma 2 implies that the measure defined by $\omega_G$ is equal to $dg$. Let U be an open subset of $(G/T)\times T_r$ such that $f_r$ induces an isomorphism from U to an open subset V of $G_r$. Let $\varphi$ be a continuous function with compact support in V; denote also by $\varphi$ the extension of $\varphi$ to $G_r$ which vanishes outside V. We have

$=\int_U(\varphi \circ f_r)\omega_{G/T}\wedge \delta_G\omega_T$ (Prop$.1$)

hence the lemma.

$\delta_G(t) =$ det(Ad$_{\mathfrak{g}/\mathfrak{t}}(t)-1$) $=\prod_{\alpha\in R(G,T)}(t^{\alpha}-1)$.

Equivalently (Integration, Chap. V, §6, no. 3, Prop. $4$)$,dg$ is the image under the map $f: (G/T)\times T\rightarrow G$ of the measure $\mu\otimes \frac{1}{w(G)}\delta_Gdt$.

We prove the last assertion. It follows from §5, no. 1 and Differentiable and Analytic Manifolds, Results$, 10.1.3c)$ that G-- $G_r$ is negligible in G and T -- $T_r$ is negligible in T. Further, the map $f_r$ makes $(G/T)\times T_r$ a principal covering of $G_r$, with group W (§5, no. 4, Prop. $4b$)$)$. The theorem now follows from Lemma 3 and Integration, Chap. V, §6, no. 6, Prop. 11.

(“Hermann Weyl’s integration formula”).

(ii) Let $\varphi$ be a positive measurable function on G. For almost all $t\in T$, the function $g \rightarrow \varphi (gtg^{-1})$ on G is measurable. The function $t \rightarrow \int_G^*\varphi (gtg^{-1})dg$ on T is measurable, and we have

Since the map $f$ is induced by passage to the quotient from the map $(g, t) \rightarrow gtg^{-1}$ from $G\times T\rightarrow G$, it suffices to apply Integration, Chap. V, §5, 6, 8 and Integration, Chap. VII, §2.

a$)\varphi$ is measurable if and only if its restriction to T is measurable.

b$)\varphi$ is integrable if and only if the function $(\varphi |T)\delta_G$ is integrable on T, and in that case we have

Denote by $p: G/T\times T\rightarrow T$ the second projection. We have $\varphi \circ f$ = $(\varphi |T)\circ p$; further, the image under $p$ of the measure $\mu\otimes \frac{1}{w(G)}\delta_Gdt$ is $\frac{1}{w(G)}\delta_Gdt$. The corollary now follows from Th. 1 above and Th. 1 of Integration, Chap. V, §6, no. 2, applied to the two proper maps $f$ and $p$.

$\int_G\varphi (g)dg=\frac{w(H)}{w(G)}\int_H\varphi (h$)det(Ad$_{\mathfrak{g}/\mathfrak{h}}(h)-1$)$dh$. (7)

Indeed, the function $h \rightarrow \varphi (h$)det(Ad$_{\mathfrak{g}/\mathfrak{h}}(h)-1$) is a central function on H whose restriction to T is the function $t \rightarrow \varphi (t)\delta_G(t)\delta_H(t)^{-1}$. Thus, the corollary follows from Cor. 2 applied to G and to H.

(exp$_H$)$^*\omega_H=\lambda_{\mathfrak{h}}\omega_{\mathfrak{h}}$ (10)

where $\lambda_{\mathfrak{h}}$ is the Ad(H)-invariant function on $\mathfrak{h}$ such that

$\lambda_{\mathfrak{h}}(x) =$ det $\sum(p+$ 1)!1 (ad $x$)$^p$ for $x\in \mathfrak{h}$.

$p\geq 0$

Let $x, x_1, . . . , x_m$ be elements of $\mathfrak{h}$. We have

(exp$^*\omega_H$)$_x(x_1, . . . , x_m) = (\omega_H$(exp $x$))$(T_x$(exp)($x_1$)$, . . . ,T_x$(exp)($x_m$)). Denote by $\varpi (x) :\mathfrak{h}\rightarrow \mathfrak{h}$ the right differential of the exponential at $x$ (Chap. III, §3, no. 17, Def. 8); by definition,

$T_x$(exp)($y$).(exp $x$)$^{-1}=\varpi (x).y$ for all $y\in \mathfrak{h}$.

The form $\omega_H$ being right invariant, we obtain

$(\omega_H$(exp $x$))$(T_x$(exp)($x_1$)$, . . . ,T_x$(exp)($x_m$))

$=\omega_H(e)(\varpi (x).x_1, . . . , \varpi (x).x_m) =$ (det $\varpi (x)$)$\omega_{\mathfrak{h}}(x_1, . . . , x_m)$;

thus, exp$^*\omega_H=\lambda_{\mathfrak{h}}\omega_{\mathfrak{h}}$, with $\lambda_{\mathfrak{h}}(x) =$ det $\varpi (x) =$ det$\frac{exp ad x-1}{ad x}$ (Chap. III, §6, no. 4, Prop. 12).

Let $h\in H$; since Ad $h$ is an automorphism of $\mathfrak{h}$, we have

ad ((Ad $h$)$(x)$) $=$ Ad $h\circ$ Ad$x\circ$ (Ad $h$)$^{-1}$,

so $\lambda_{\mathfrak{h}}$((Ad $h$)$(x)$) $=\lambda_{\mathfrak{h}}(x)$. Thus, the function $\lambda_{\mathfrak{h}}$ is invariant under Ad(H); this completes the proof of the proposition.

Denote by $\omega_G$ (resp. $\omega_T$) the invariant differential form of maximum degree on G (resp. T) that coincides with $\omega_{\mathfrak{g}}$ (resp. $\omega_{\mathfrak{t}}$) at the origin. Consider the commutative diagram

(Id,exp$_T$) exp$_G$.

$(G/T)\times T\longrightarrow^f$ G

In view of Prop. 1 of no. 2 and the relation exp$^*_T\omega_T=\omega_{\mathfrak{t}}$, we deduce the equality

$\psi^*$exp$^*_G\omega_G=\omega_{G/T}\wedge \delta_{\mathfrak{g}}\omega_{\mathfrak{t}}$.

By Prop. $2,\psi^*$exp$^*_G\omega_G= (\psi^*\lambda_{\mathfrak{g}})\psi^*\omega_{\mathfrak{g}}$. Since the function $\lambda_{\mathfrak{g}}$ is invariant under Ad(G), we have $(\psi^*\lambda_{\mathfrak{g}})(\overline{g}, x) =\lambda_{\mathfrak{g}}(x) =\frac{\delta_{\mathfrak{g}}(x)}{\pi_{\mathfrak{g}}(x)}$ for $\overline{g}\in G/T, x\in \mathfrak{t}$.

It follows that the forms $\psi^*\omega_G(\overline{g}, x)$ and $\omega_{G/T}(\overline{g})\wedge \pi_{\mathfrak{g}}(x)\omega_{\mathfrak{t}}(x)$ coincide where $\delta_{\mathfrak{g}}(x)$ is non-zero, that is on the dense open subset $(G/T)\times \mathfrak{t}_r$; thus, they are equal, hence the proposition.

Choose invariant differential forms $\omega_G$ on G and $\omega_T$ on T, of maximum degree, such that $|\omega_G|=dg$ and $|\omega_T|=dt$; denote by $\omega_{\mathfrak{g}}$ (resp. $\omega_{\mathfrak{t}}$) the translation-invariant differential form on $\mathfrak{g}$ (resp. $\mathfrak{t}$) that coincides with $\omega_G(e)$ (resp. $\omega_T(e)$) at the origin, and $dz$ (resp. $dx$) the Haar measure $|\omega_{\mathfrak{g}}|$ (resp. $|\omega_{\mathfrak{t}}|$). Reasoning as in no. 2, mutatis mutandis, gives the following proposition:

We leave to the reader the statement and proof of the analogues of Cor. 1 to 3 and Remarks 1 to 3 of no. 2. For example, let $\varphi$ be an integrable function on $\mathfrak{g}$ (with values in a Banach space or $\overline{\mathbf{R}}$); then

$\int_{\mathfrak{g}}\varphi (z)dz=\frac{1}{w(G)}\int_{\mathfrak{t}}(\int_G\varphi$((Ad $g$)$x$)$dg)\pi_{\mathfrak{g}}(x)dx$, (12)

and, in particular, if $\varphi$ is invariant under Ad(G),