Orthogonal Matrix Integration

H1H_1H1 is an n×mn\times mn×m matrix ( n>mn > mn>m ) with orthonormal columns, so that H1′H1=ImH_1'H_1 = I_mH1′H1=Im. The set (or space) of all such matrices H1H_1H1 is called the Stiefel manfold, denoted by Vm,nV_{m,n}Vm,n.Thus
Vm,n={H1(n×m):H1′H1=Im}V_{m,n} =\{H_1(n\times m):H_1'H_1 = I_m\}Vm,n={H1(n×m):H1′H1=Im}
There are 12m(m+1)\frac{1}{2} m ( m + 1)21m(m+1) functionally independent conditions on the mnmnmn elements of H1∈Vm,nH_1\in V_{m,n}H1∈Vm,n implied by the equation H1′H1=ImH_1'H_1 = I_mH1′H1=Im. Hence the elements of H1H_1H1 can be regarded as the coordinates of a point on a mn−12m(m+1)mn - \frac{1}{2} m ( m + 1)mn−21m(m+1)-dimensional surface in mn-dimensional Euclidean space.If H1=(hi,j)(i=1,...,n;j=1,...,m)H_1 = ( h_{i,j}) ( i =1 , . . . ,n; j = 1 , . . . ,m)H1=(hi,j)(i=1,...,n;j=1,...,m) then since ∑i=1n∑j=1mhi,j2=m\sum _{i=1}^n\sum_{j=1}^m h_{i,j}^2 = m∑i=1n∑j=1mhi,j2=m this surface is a subset of the sphere of radius m1/2m^{1/2}m1/2 in mn-dimensional space.

The 12m(m+1)\frac{1}{2} m ( m + 1)21m(m+1) functionally independent conditions:
<hj,hj>=1,(j=1,...,m)<hi,hj>=0,(i>j)\begin {aligned} &<h_j,h_j>=1, (j = 1 , . . . ,m)\\ &<h_i,h_j>=0, (i>j) \end{aligned}<hj,hj>=1,(j=1,...,m)<hi,hj>=0,(i>j)

Two special cases are the following:

a) m=nm=nm=n
Vm,m≡O(m)={H1(m×m):H1′H1=Im}V_{m,m}\equiv O(m)=\{H_1(m\times m):H_1'H_1 = I_m\}Vm,m≡O(m)={H1(m×m):H1′H1=Im}
Here the elements of O(m)O(m)O(m) can be regarded as the coordinates of a point on a 12m(m−1)\frac{1}{2} m ( m - 1)21m(m−1)-dimensional surface in Euclidean m2m^2m2-space and the surface is a subset of the sphere of radius m1/2m^{1/2}m1/2 in m2m^2m2-space.

a) m=1m=1m=1
V1,n≡Sn={H1(n×1):H1′H1=1}V_{1,n}\equiv S_n=\{H_1(n\times 1):H_1'H_1 = 1\}V1,n≡Sn={H1(n×1):H1′H1=1}
the unit sphere in RnR^nRn. This is, of course, an n−1n - 1n−1 dimensional surface in RnR^nRn.

THEOREM 1
∫Vm,nH1′dH1=2mπmn/2Γm(12n)(1)\int_{{\rm {\bf V_{m,n}}} } {{\rm {\bf H_1'dH_1}}} = \frac{2^m\pi ^{mn / 2}}{\Gamma _m (\frac{1}{2}n)} \tag 1∫Vm,nH1′dH1=Γm(21n)2mπmn/2(1)

Proof: Let ZZZ be an n×m,(n≥m)n\times m,(n\ge m)n×m,(n≥m) random matrix whose elements are
all independent N(0,1)N(0,1)N(0,1) random variables. The density function of ZZZ (that is,
the joint density function of the mn elements of ZZZ ) is

f(Z)=(2π)−mn/2exp(−12∑i=1n∑j=1mZij2)=(2π)−mn/2etr(−12Z′Z)(2)\begin {aligned} f({\rm {\bf Z}}) &={\left( {2\pi } \right)^{-mn / 2}}\mathrm{exp}\left( - \frac{1}{2}\sum_{i=1}^n\sum_{j=1}^m Z_{ij}^2 \right)\\ &={\left( {2\pi } \right)^{-mn / 2}}\mathrm{etr}\left( { - \frac{1}{2}{\rm {\bf Z'Z}}} \right) \end{aligned} \tag 2f(Z)=(2π)−mn/2exp(−21i=1∑nj=1∑mZij2)=(2π)−mn/2etr(−21Z′Z)(2)
It is very known that
∫Rn×mf(Z)(dZ)=1\int_{R^{n\times m}} f({\rm {\bf Z}})(d{\bf Z})=1∫Rn×mf(Z)(dZ)=1
That is
∫⋯∫Zij∈(−∞,∞)etr(−12Z′Z)(dZ)=(2π)mn/2{\int\cdots\int}_{Z_{ij}\in(-\infty,\infty)} \mathrm{etr}\left( { - \frac{1}{2}{\rm {\bf Z'Z}}} \right)(d{\bf Z})={\left( {2\pi } \right)^{mn / 2}}∫⋯∫Zij∈(−∞,∞)etr(−21Z′Z)(dZ)=(2π)mn/2

where (dZ)=∏ijdZij,(i=1,...,n,j=1,...,m)(d{\bf Z})=\prod_{ij}d Z_{ij},(i=1,...,n, j=1,...,m)(dZ)=∏ijdZij,(i=1,...,n,j=1,...,m).

Put Z=H1T{\bf Z} = H_1 TZ=H1T, where H1∈Vm,nH_1\in V_{m,n}H1∈Vm,n and TTT is upper-triangular with positive diagonal elements, then from 2.13 of ref1
(dZ)=∏i=1mtiin−i(dT)(H1′dH1)(3)\left( {d{\rm {\bf Z}}} \right) = \prod\limits_{i = 1}^m {t_{ii}^{n - i} } \left( {dT} \right)\left( {H_1'dH_1} \right) \tag 3(dZ)=i=1∏mtiin−i(dT)(H1′dH1)(3)

∫etr(−12Z′Z)(dZ)=∫etr(−12T′T)∏i=1mtiin−i(dT)∫(H1′dH1)\int \mathrm{etr}\left( { - \frac{1}{2}{\rm {\bf Z'Z}}} \right)(d{\bf Z})= \int \mathrm{etr}\left( { - \frac{1}{2}{\rm {\bf T'T}}} \right)\prod\limits_{i = 1}^m {t_{ii}^{n - i} } \left( {dT} \right)\int\left( {H_1'dH_1} \right)∫etr(−21Z′Z)(dZ)=∫etr(−21T′T)i=1∏mtiin−i(dT)∫(H1′dH1)

∫etr(−12T′T)∏i=1mtiin−i(dT)=∏i<j,j=2mexp⁡(−12tij2)dtij∫∏i=1mexp⁡(−12tii2)tiin−idtii=(2π)m(m−1)/4∏i=1m∫si>0exp⁡(−si)(2si)(n−i−1)/2dsi=πm(m−1)/4∏i=1mΓ[n−i+12]2m(m−1)/4∏i=1m2(n−i−1)/2=Γm(n2)⋅2mn/2−m.\begin {aligned} &\int \mathrm{etr}\left( { - \frac{1}{2}{\rm {\bf T'T}}} \right)\prod\limits_{i = 1}^m {t_{ii}^{n - i} } \left( {dT} \right)\\&= \prod_{i<j,j=2}^m\exp(-\frac{1}{2}t_{ij}^2)dt_{ij}\int \prod_{i=1}^m\exp(-\frac{1}{2}t_{ii}^2)t_{ii}^{n - i}dt_{ii}\\ &=(2\pi)^{m(m-1)/4} \prod_{i=1}^m\int_{s_i>0}\exp(-s_i)(2s_i)^{(n-i-1)/2}ds_i\\ &=\pi^{m(m-1)/4} \prod_{i=1}^m\Gamma \left [\frac{n-i+1}{2}\right ]2^{m(m-1)/4}\prod_{i=1}^m 2^{(n-i-1)/2}\\ &=\Gamma_m(\frac{n}{2})\cdot 2^{mn/2-m}. \end{aligned}∫etr(−21T′T)i=1∏mtiin−i(dT)=i<j,j=2∏mexp(−21tij2)dtij∫i=1∏mexp(−21tii2)tiin−idtii=(2π)m(m−1)/4i=1∏m∫si>0exp(−si)(2si)(n−i−1)/2dsi=πm(m−1)/4i=1∏mΓ[2n−i+1]2m(m−1)/4i=1∏m2(n−i−1)/2=Γm(2n)⋅2mn/2−m.
Then
∫Vm,nH1′dH1=2mπmn/2Γm(12n)\int_{{\rm {\bf V_{m,n}}} } {{\rm {\bf H_1'dH_1}}} = \frac{2^m\pi ^{mn / 2}}{\Gamma _m (\frac{1}{2}n)} ∫Vm,nH1′dH1=Γm(21n)2mπmn/2
End of proof. ♢\diamondsuit♢

Here
Γm(n2)=πm(m−1)/4∏i=1mΓ[n−i+12](4)\Gamma_m(\frac{n}{2})=\pi^{m(m-1)/4} \prod_{i=1}^m\Gamma \left [\frac{n-i+1}{2}\right ]\tag 4 Γm(2n)=πm(m−1)/4i=1∏mΓ[2n−i+1](4)
Like Γ(n)=∫x>0e−xxndx\Gamma(n)=\int_{x>0}e^{-x}x^n dxΓ(n)=∫x>0e−xxndx
Γm(n)=∫A>0etr(−A)(det⁡A)(n−(m+1)/2)dA(5)\Gamma_m(n)=\int_{A>0} \mathrm {etr}(-A)(\det A)^{(n-(m+1)/2)}dA\tag 5 Γm(n)=∫A>0etr(−A)(detA)(n−(m+1)/2)dA(5)
where A is a positive m×mm\times mm×m matrix. Using A=T′TA=T'TA=T′T and
dA=2m∏i=1mtiim−i+1(dT)(6)dA=2^m\prod_{i=1}^m t_{ii}^{m-i+1}(dT)\tag 6dA=2mi=1∏mtiim−i+1(dT)(6)

Γm(n)=∏i<j,j=2mexp⁡(−tij2)dtij∏i=1m∫2exp⁡(−tii2)tii2n−idtii=πm(m−1)/4∏i=1m∫ui>0exp⁡(−ui)uin−i/2−1/2dui=πm(m−1)/4∏i=1mΓ(n−i/2+1/2)\begin{aligned} \Gamma_m(n)&= \prod_{i<j,j=2}^m\exp(-t_{ij}^2)dt_{ij} \prod_{i=1}^m\int 2\exp(-t_{ii}^2)t_{ii}^{2n - i}dt_{ii}\\ &=\pi^{m(m-1)/4}\prod_{i=1}^m\int_{u_i>0} \exp(-u_{i})u_{i}^{n - i/2-1/2}du_{i}\\ &=\pi^{m(m-1)/4}\prod_{i=1}^m\Gamma(n-i/2+1/2) \end{aligned}Γm(n)=i<j,j=2∏mexp(−tij2)dtiji=1∏m∫2exp(−tii2)tii2n−idtii=πm(m−1)/4i=1∏m∫ui>0exp(−ui)uin−i/2−1/2dui=πm(m−1)/4i=1∏mΓ(n−i/2+1/2)

THEOREM 1C (for complex)
∫Vm,nH1HdH1=2mπmnΓ~m(n)(7)\int_{{\rm {\bf V_{m,n}}} } {{\rm {\bf H_1^HdH_1}}} = \frac{2^m\pi ^{mn}}{\widetilde{\Gamma}_m (n)}\tag 7 ∫Vm,nH1HdH1=Γm(n)2mπmn(7)

Proof: Let ZZZ be an n×m,(n≥m)n\times m,(n\ge m)n×m,(n≥m) random matrix whose elements are
all independent N(0,1)N(0,1)N(0,1) random complex variables with real and imagine part of N(0,1/2)N(0,1/2)N(0,1/2) . The density function of ZZZ (that is, the joint density function of the mn elements of ZZZ ) is

f(Z)=(π)−mnexp(−∑i=1n∑j=1mZij∗Zij)=(π)−mnetr(−ZHZ)(8)\begin {aligned} f({\rm {\bf Z}}) &={\left( {\pi } \right)^{-mn}} \mathrm{exp}\left( - \sum_{i=1}^n\sum_{j=1}^m Z_{ij}^*Z_{ij} \right)\\ &={\left( {\pi } \right)^{-mn}} \mathrm{etr}\left( { - { {\bf Z^HZ}}} \right) \end{aligned}\tag 8f(Z)=(π)−mnexp(−i=1∑nj=1∑mZij∗Zij)=(π)−mnetr(−ZHZ)(8)
It is very known that
∫Cn×mf(Z)(dZ)=1\int_{\mathcal{C}^{n\times m}} f({\rm {\bf Z}})(d{\bf Z})=1 ∫Cn×mf(Z)(dZ)=1
That is
∫⋯∫Zijetr(−ZHZ)(dZ)=(π)mn{\int\cdots\int}_{Z_{ij}} \mathrm{etr}\left( { - {\rm {\bf Z^HZ}}} \right)(d{\bf Z})={\left( {\pi } \right)^{mn }} ∫⋯∫Zijetr(−ZHZ)(dZ)=(π)mn
where (dZ)=⋀ijdZij,(i=1,...,n,j=1,...,m)(d{\bf Z})=\bigwedge_{ij}d Z_{ij},(i=1,...,n, j=1,...,m)(dZ)=⋀ijdZij,(i=1,...,n,j=1,...,m).

Put Z=H1T{\bf Z} = H_1 TZ=H1T, where H1∈Vm,nH_1\in V_{m,n}H1∈Vm,n and TTT is upper-triangular with positive diagonal elements, then from 2.13 of ref1
(dZ)=∏i=1mtii2n−2i+1(dT)(H1HdH1)(9)\left( {d{\rm {\bf Z}}} \right) = \prod\limits_{i = 1}^m {t_{ii}^{2n - 2i+1} } \left( {dT} \right)\left( {H_1^HdH_1} \right)\tag 9 (dZ)=i=1∏mtii2n−2i+1(dT)(H1HdH1)(9)

∫etr(−ZHZ)(dZ)=∫etr(−THT)∏i=1mtii2n−2i+1(dT)∫(H1HdH1)\int \mathrm{etr}\left( { - {\rm {\bf Z^HZ}}} \right)(d{\bf Z})= \int \mathrm{etr}\left( { - {\rm {\bf T^HT}}} \right)\prod\limits_{i = 1}^m {t_{ii}^{2n - 2i+1} } \left( {dT} \right)\int\left( {H_1^HdH_1} \right) ∫etr(−ZHZ)(dZ)=∫etr(−THT)i=1∏mtii2n−2i+1(dT)∫(H1HdH1)

∫etr(−THT)∏i=1mtii2n−2i+1(dT)=∏i<j,i=1,j=2mexp⁡(−ℜ(tij)2)dℜ(tij)exp⁡(−ℑ(tij)2)dℑ(tij)∫∏i=1mexp⁡(−tii2)tii2n−2i+1dtii=(π)m(m−1)/2∏i=1m∫si>02−1exp⁡(−si)(si)n−idsi=πm(m−1)/2∏i=1mΓ[n−i+1]2−m=Γ~m(n).2−m.\begin {aligned} &\int \mathrm{etr}\left( { - {\rm {\bf T^HT}}} \right)\prod\limits_{i = 1}^m {t_{ii}^{2n - 2i+1} } \left( {dT} \right)\\&= \prod_{i<j,i=1,j=2}^m\exp(-\Re(t_{ij})^2)d\Re(t_{ij})\exp(-\Im(t_{ij})^2)d\Im(t_{ij})\int \prod_{i=1}^m\exp(-t_{ii}^2)t_{ii}^{2n - 2i+1}dt_{ii}\\ &=(\pi)^{m(m-1)/2} \prod_{i=1}^m\int_{s_i>0}2^{-1}\exp(-s_i)(s_i)^{n-i}ds_i\\ &=\pi^{m(m-1)/2} \prod_{i=1}^m\Gamma \left [{n-i+1}\right ]2^{-m}\\ &=\widetilde{\Gamma}_m(n).2^{-m}. \end{aligned}∫etr(−THT)i=1∏mtii2n−2i+1(dT)=i<j,i=1,j=2∏mexp(−ℜ(tij)2)dℜ(tij)exp(−ℑ(tij)2)dℑ(tij)∫i=1∏mexp(−tii2)tii2n−2i+1dtii=(π)m(m−1)/2i=1∏m∫si>02−1exp(−si)(si)n−idsi=πm(m−1)/2i=1∏mΓ[n−i+1]2−m=Γm(n).2−m.
Note: Here tij,i<jt_{ij},i<jtij,i<j are complex, tiit_{ii}tii are real.
Then
∫Vm,nH1HdH1=2mπmnΓ~m(n)\int_{{\rm {\bf V_{m,n}}} } {{\rm {\bf H_1^HdH_1}}} = \frac{2^m\pi ^{mn}}{\widetilde{\Gamma} _m (n)} ∫Vm,nH1HdH1=Γm(n)2mπmn
End of proof. ♢♢\diamondsuit \diamondsuit♢♢

Here
Γ~m(n)=πm(m−1)/2∏i=1mΓ[n−i+1](10)\widetilde{\Gamma}_m(n)=\pi^{m(m-1)/2} \prod_{i=1}^m\Gamma \left [{n-i+1}\right ]\tag {10} Γm(n)=πm(m−1)/2i=1∏mΓ[n−i+1](10)
Like Γ(n)=∫x>0e−xxndx\Gamma(n)=\int_{x>0}e^{-x}x^n dxΓ(n)=∫x>0e−xxndx
Γ~m(n)=∫A>0etr(−A)(det⁡A)n−mdA(11)\begin{aligned} \widetilde{\Gamma}_m(n)=\int_{A>0}\mathrm {etr}(-A)(\det A)^{n-m}dA \end{aligned} \tag {11}Γm(n)=∫A>0etr(−A)(detA)n−mdA(11)
where A is a positive m×mm\times mm×m complex matrix. Using A=T′TA=T'TA=T′T and
dA=2m∏i=1mtii2m−2i+1(dT)det⁡A=∏i=1mtii2(12)\begin{aligned} dA=2^m\prod_{i=1}^m t_{ii}^{2m-2i+1}(dT)\\ \det A=\prod_{i=1}^m t_{ii}^2 \tag {12} \end{aligned}dA=2mi=1∏mtii2m−2i+1(dT)detA=i=1∏mtii2(12)

Γ~m(n)=∏i<j,i=1,j=2mexp⁡(−∣tij∣2)dℜ(tij)dℑ(tij)∏i=1m∫2exp⁡(−tii2)tii2n−2i+1dtii=πm(m−1)/2∏i=1m∫ui>0exp⁡(−ui)uin−idui=πm(m−1)/2∏i=1mΓ(n−i+1)\begin{aligned} \widetilde{\Gamma}_m(n)&= \prod_{i<j,i=1,j=2}^m\exp(-|t_{ij}|^2)d\Re(t_{ij})d\Im(t_{ij}) \prod_{i=1}^m\int 2\exp(-t_{ii}^2)t_{ii}^{2n - 2i+1}dt_{ii}\\ &=\pi^{m(m-1)/2}\prod_{i=1}^m\int_{u_i>0} \exp(-u_{i})u_{i}^{n - i}du_{i} \\ &=\pi^{m(m-1)/2}\prod_{i=1}^m\Gamma(n-i+1) \end{aligned}Γm(n)=i<j,i=1,j=2∏mexp(−∣tij∣2)dℜ(tij)dℑ(tij)i=1∏m∫2exp(−tii2)tii2n−2i+1dtii=πm(m−1)/2i=1∏m∫ui>0exp(−ui)uin−idui=πm(m−1)/2i=1∏mΓ(n−i+1)

Derivation of (6): for A=T′TA=T'TA=T′T.
a11=t112da11=2t11dt11a21=t12t11da21=t11dt12+∗a31=t13t11da31=t11dt13+∗⋯⋯am1=t1mt11dam1=t11dt1m+∗a22=t122+t222da22=2t22dt22+∗a32=t13t12+t23t22da32=t22dt23+∗⋯⋯am2=t1mt12+t2mt22dam2=t22dt2m+∗⋯⋯amm=tmm2+∗damm=2tmmdtmm+∗\begin{aligned} \begin{array}{ll} a_{11}=t_{11}^2 & da_{11}=2t_{11}dt_{11}\\ a_{21}=t_{12}t_{11} & da_{21}=t_{11}dt_{12}+*\\ a_{31}=t_{13}t_{11} & da_{31}=t_{11}dt_{13}+*\\ \cdots &\cdots\\ a_{m1}=t_{1m}t_{11} & da_{m1}=t_{11}dt_{1m}+*\\ a_{22}=t_{12}^2+t_{22}^2 & da_{22}=2t_{22}dt_{22}+*\\ a_{32}=t_{13}t_{12}+ t_{23}t_{22}& da_{32}=t_{22}dt_{23}+*\\ \cdots &\cdots\\ a_{m2}=t_{1m}t_{12}+ t_{2m}t_{22}& da_{m2}=t_{22}dt_{2m}+*\\ \cdots &\cdots\\ a_{mm}=t_{mm}^2+* & da_{mm}=2t_{mm}dt_{mm}+* \end{array} \end{aligned}a11=t112a21=t12t11a31=t13t11⋯am1=t1mt11a22=t122+t222a32=t13t12+t23t22⋯am2=t1mt12+t2mt22⋯amm=tmm2+∗da11=2t11dt11da21=t11dt12+∗da31=t11dt13+∗⋯dam1=t11dt1m+∗da22=2t22dt22+∗da32=t22dt23+∗⋯dam2=t22dt2m+∗⋯damm=2tmmdtmm+∗
dA=2m∏i=1mtiim−i+1⋀i≤j,j=1mdtij\begin{aligned} dA=2^m\prod_{i=1}^m t_{ii}^{m-i+1}\bigwedge_{i\le j,j=1}^m dt_{ij}\end{aligned}dA=2mi=1∏mtiim−i+1i≤j,j=1⋀mdtij
Derivation of (12):
For a complex matrix A=THTA=T^HTA=THT,
aiia_{ii}aii are real, daii=2tiidtii+∗da_{ii}=2t_{ii}dt_{ii}+*daii=2tiidtii+∗,
aij=xij+yij,i>ja_{ij}=x_{ij}+y_{ij}, i> jaij=xij+yij,i>j are complex,including real part xijx_{ij}xij and imagine part yijy_{ij}yij,
dxij=tiidℜ(tji)+∗,dyij=−tiidℑ(tji)+∗\begin{aligned}dx_{ij}=t_{ii}d\Re({t}_{ji})+*,\\dy_{ij}=-t_{ii}d\Im({t}_{ji})+* \end{aligned}dxij=tiidℜ(tji)+∗,dyij=−tiidℑ(tji)+∗
So
dA=2m∏i=1mtii2m−2i+1⋀i≤j,j=1mdtij\begin{aligned} dA=2^m\prod_{i=1}^m t_{ii}^{2m-2i+1}\bigwedge_{i\le j,j=1}^m dt_{ij}\end{aligned}dA=2mi=1∏mtii2m−2i+1i≤j,j=1⋀mdtij

THEOREM 2.1.13. Let ZZZ be an n×m,(n≥m)n\times m, ( n \ge m )n×m,(n≥m) matrix of rank m and
write Z=H1TZ= H_1TZ=H1T, where H1H_1H1 is an n×mn\times mn×m matrix with H1′H1=ImH_1'H_1=I_mH1′H1=Im, and T is an m×mm\times mm×m upper-triangular matrix with positive diagonal elements. Let H2H_2H2 (a function of H1H_1H1 ) be an n×(n−m)n \times ( n - m )n×(n−m) matrix such that H=[H1:H2)H = [ H_1 : H_2)H=[H1:H2)is an orthogonal n×nn \times nn×n matrix and write H=[h1,...hm:hm+1,...hn]H=[h_1,. .. h_m: h_{m+1},. . . h_n]H=[h1,...hm:hm+1,...hn], where h1,...hmh_1,. .. h_mh1,...hm are the columns of H1H_1H1 and hm+1,...hnh_{m+1},. . . h_nhm+1,...hn are the columns of H2H_2H2. Then
(dZ)=∏i=1mtiin−i(dT)(H1′dH1)(13)\begin{aligned} \left( {d{\rm {\bf Z}}} \right) = \prod\limits_{i = 1}^m {t_{ii}^{n - i} } \left( {dT} \right)\left( {H_1'dH_1} \right) \end{aligned} \tag {13}(dZ)=i=1∏mtiin−i(dT)(H1′dH1)(13)

Proof. Z=H1T⟹dZ=dH1.T+H1.dTZ= H_1T\Longrightarrow dZ=dH_1.T+H_1.dTZ=H1T⟹dZ=dH1.T+H1.dT hence
H′dZ=[H1′H2′¨]dZ=[H1′dH1T+H1′H1dTH2′dH1T+H2′H1dT]=[H1′dH1T+dTH2′dH1T]\begin {aligned} H'{d{\bf Z}}&= \begin{bmatrix} H_1'\\ \ddot{H_2'} \end{bmatrix} d {\bf Z}=\begin{bmatrix} H_1'dH_1T+H_1'H_1dT\\ H_2'dH_1T+H_2'H_1dT \end{bmatrix} \\ &=\begin{bmatrix} H_1'dH_1T+dT\\ H_2'dH_1T \end{bmatrix} \end{aligned}H′dZ=[H1′H2′¨]dZ=[H1′dH1T+H1′H1dTH2′dH1T+H2′H1dT]=[H1′dH1T+dTH2′dH1T]

since H1′H1=Im,H1′H2=0H_1'H_1=I_m, H_1'H_2=0H1′H1=Im,H1′H2=0. By Theorem 2.1.4 the exterior product of the elements on the left side of (13) is
(H′dZ)=(det⁡H′)m(dZ)=(dZ)(13.1)\begin {aligned} \left( H'{d{\rm {\bf Z}}} \right) = (\det H')^m (d{\bf Z})=(d{\bf Z}) \end{aligned}\tag {13.1}(H′dZ)=(detH′)m(dZ)=(dZ)(13.1)
Proof of (13.1).
(dZ)=⋀j=1m⋀i=1ndZij=⋀j=1m(dZj)\begin {aligned} \left( {d{\rm {\bf Z}}} \right) =\bigwedge_{j=1}^m\bigwedge_{i=1}^n dZ_{ij}=\bigwedge_{j=1}^m (dZ_{j}) \end{aligned}(dZ)=j=1⋀mi=1⋀ndZij=j=1⋀m(dZj)
Considering exterior product of the same element dxΛdx=0dx\Lambda dx=0dxΛdx=0, here dZijΛdZij=0dZ_{ij}\Lambda dZ_{ij}=0dZijΛdZij=0, and the definition of determinant (the algebraic sum of products of different rows and column of the matrix),

(H′dZj)=(h11dZ1j+h21dZ2j+⋯+hn1dZnjh12dZ1j+h22dZ2j+⋯+hn2dZnj⋯h1ndZ1j+h2ndZ2j+⋯+hnndZnj)=det⁡(H)⋀i=1ndZij=det⁡(H)(dZj)\begin {aligned} (H'dZ_{j}) &=\begin{pmatrix} h_{11}dZ_{1j}+h_{21}dZ_{2j}+\cdots+h_{n1}dZ_{nj}\\ h_{12}dZ_{1j}+h_{22}dZ_{2j}+\cdots+h_{n2}dZ_{nj}\\ \cdots\\ h_{1n}dZ_{1j}+h_{2n}dZ_{2j}+\cdots+h_{nn}dZ_{nj}\\ \end{pmatrix}\\ &=\det(H)\bigwedge_{i=1}^n dZ_{ij}=\det(H)( dZ_{j}) \end{aligned}(H′dZj)=⎝⎜⎜⎛h11dZ1j+h21dZ2j+⋯+hn1dZnjh12dZ1j+h22dZ2j+⋯+hn2dZnj⋯h1ndZ1j+h2ndZ2j+⋯+hnndZnj⎠⎟⎟⎞=det(H)i=1⋀ndZij=det(H)(dZj)

(H′dZ)=⋀j=1m(H′dZj)=det⁡(H)m(dZ)=(dZ)\begin{aligned} \left( H'{d{\rm {\bf Z}}} \right) =\bigwedge_{j=1}^m (H'dZ_{j})=\det(H)^m(d\rm{Z})=(d\rm{Z}) \end{aligned}(H′dZ)=j=1⋀m(H′dZj)=det(H)m(dZ)=(dZ)
for det⁡(H)=1\det (H)=1det(H)=1.
End of proof of (13.1).

First consider the matrix H2′dH1TH_2' dH_1 TH2′dH1T. The (j−m)(j- m)(j−m)th row of H2′dH1TH_2' dH_1 TH2′dH1T is
(hj′dh1,⋯,,,...,hj′dhm)T,(m+1≤j≤n).\begin{aligned}\begin{matrix} (h_j'dh_1,\cdots, ,,..., h_j'dh_m)T, &&&( m + 1 \le j \le n ) . \end{matrix}\end{aligned}(hj′dh1,⋯,,,...,hj′dhm)T,(m+1≤j≤n).
((hj′dh1,⋯,,,...,hj′dhm)T)=(det⁡T)⋀i=1mhj′dhi.\begin{aligned} ((h_j'dh_1,\cdots, ,,..., h_j'dh_m)T)=(\det T)\bigwedge_{i=1}^m h_j'dh_i. \end{aligned}((hj′dh1,⋯,,,...,hj′dhm)T)=(detT)i=1⋀mhj′dhi.
so
(H2′dH1T)=(det⁡T)n−m⋀j=m+1n⋀i=1mhj′dhi.(13.2)\begin{aligned} (H_2' dH_1 T)=(\det T)^{n-m} \bigwedge_{j=m+1}^n \bigwedge_{i=1}^m h_j'dh_i. \end{aligned}\tag{13.2}(H2′dH1T)=(detT)n−mj=m+1⋀ni=1⋀mhj′dhi.(13.2)

Now consider the upper matrix H1′dH1T+dTH_1' dH_1 T+dTH1′dH1T+dT. First note that since H1′H1=ImH_1' H_1=I_mH1′H1=Im, we have
H1′dH1+dH1′.H1=0\begin{aligned} H_1' dH_1+dH_1'.H_1=0 \end{aligned}H1′dH1+dH1′.H1=0
so that
H1′dH1=−dH1′.H1=−(H1′dH1)′,(13.2)\begin{aligned} H_1' dH_1=-dH_1'.H_1=-(H_1' dH_1)', \end{aligned}\tag {13.2}H1′dH1=−dH1′.H1=−(H1′dH1)′,(13.2)
and hence H1′dH1H_1' dH_1H1′dH1, is skew-symmetric:
H1′dH1=[0−h2′dh1⋯−hm′dh1h2′dh10⋯−hm′dh2h3′dh1h3′dh2⋯−hm′dh3⋮⋮⋮hm′dh1hm′dh20](13.3)\begin{aligned} H_1' dH_1=\begin{bmatrix} 0 & -h_2'dh_1 & \cdots & -h_m'dh_1\\ h_2'dh_1 & 0 & \cdots & -h_m'dh_2\\ h_3'dh_1 & h_3'dh_2 & \cdots & -h_m'dh_3\\ \vdots & \vdots & &\vdots\\ h_m'dh_1 & h_m'dh_2 & &0 \end{bmatrix} \end{aligned}\tag{13.3}H1′dH1=⎣⎢⎢⎢⎢⎢⎡0h2′dh1h3′dh1⋮hm′dh1−h2′dh10h3′dh2⋮hm′dh2⋯⋯⋯−hm′dh1−hm′dh2−hm′dh3⋮0⎦⎥⎥⎥⎥⎥⎤(13.3)

Postmultiplying this by the upper-triangular matrix T gives the following matrix, where only the subdiagonal elements are given, and where, in addition, terms of the form hi′dhjh_i' d h_jhi′dhj are ignored if they have appeared already in a previous column:
(My comment: hi′dhjh_i' d h_jhi′dhj can be considered as a direction, the exterior product of the same directional vector is 0.)
H1′dH1T=[0∗⋯∗∗h2′dh1t11∗⋯∗∗h3′dh1t11h3′dh2t22+∗⋯∗∗⋮⋮⋮hm′dh1t11hm′dh2t22+∗−hm′dhmtm−1,m−1+∗∗](13.4)\begin{aligned} H_1' dH_1T=\begin{bmatrix} 0 & * & \cdots & *& *\\ h_2'dh_1t_{11} & * & \cdots & *& *\\ h_3'dh_1t_{11} & h_3'dh_2t_{22}+* & \cdots & *& *\\ \vdots & \vdots & &\vdots\\ h_m'dh_1t_{11} & h_m'dh_2t_{22}+* & &-h_m'dh_mt_{m-1,m-1}+*& * \end{bmatrix} \end{aligned}\tag{13.4}H1′dH1T=⎣⎢⎢⎢⎢⎢⎡0h2′dh1t11h3′dh1t11⋮hm′dh1t11∗∗h3′dh2t22+∗⋮hm′dh2t22+∗⋯⋯⋯∗∗∗⋮−hm′dhmtm−1,m−1+∗∗∗∗∗⎦⎥⎥⎥⎥⎥⎤(13.4)

Column by column, the exterior product of the subdiagonal elements (under the diagonal)of H1′dH1T+dTH_1' dH_1 T+dTH1′dH1T+dT is (remember that dTdTdT is upper-triangular)

t11m−1t22m−2⋯tm−1,m−1⋀i=1m⋀j=i+1mhj′dhi.\begin{aligned} t_{11}^{m-1}t_{22}^{m-2}\cdots t_{m-1,m-1}\bigwedge_{i=1}^m\bigwedge_{j=i+1}^m h_j'dh_i. \end{aligned}t11m−1t22m−2⋯tm−1,m−1i=1⋀mj=i+1⋀mhj′dhi.
(H1′dH1T+dT)=t11m−1t22m−2⋯tm−1,m−1⋀i=1m⋀j=i+1mhj′dhi(dT).\begin{aligned} (H_1' dH_1T+dT)=t_{11}^{m-1}t_{22}^{m-2}\cdots t_{m-1,m-1}\bigwedge_{i=1}^m\bigwedge_{j=i+1}^m h_j'dh_i(dT). \end{aligned}(H1′dH1T+dT)=t11m−1t22m−2⋯tm−1,m−1i=1⋀mj=i+1⋀mhj′dhi(dT).

The exterior product of the elements of H2′dH1TH_2' dH_1 TH2′dH1T
and the subdiagonal elements of H1′dH1T+dTH_1' dH_1 T+dTH1′dH1T+dT is
∏i=1mtiin−i⋀i=1m⋀j=i+1nhj′dhi=∏i=1mtiin−i(H1′dH1).(13.5)\begin{aligned} \prod_{i=1}^m t_{ii}^{n-i}\bigwedge_{i=1}^m\bigwedge_{j=i+1}^n h_j'dh_i=\prod_{i=1}^m t_{ii}^{n-i}(H_1'dH_1). \end{aligned}\tag{13.5}i=1∏mtiin−ii=1⋀mj=i+1⋀nhj′dhi=i=1∏mtiin−i(H1′dH1).(13.5)

The exterior product of the elements of $ H_1’ dH_1 T+dT$ on and
above the diagonal is
⋀i≤jm+terms involving dH1.\begin{aligned} \bigwedge_{i\le j}^m + \textrm {terms involving }{d}H_1. \end{aligned}i≤j⋀m+terms involving dH1.
The terms of dH1{d}H_1dH1 do not contribute to the total exterior product.Hence the exterior product of the elements of the right side of (13) is
∏i=1mtiin−i⋀i=1m⋀j=i+1nhj′dhi=∏i=1mtiin−i(dT)(H1′dH1).\begin{aligned} \prod_{i=1}^m t_{ii}^{n-i}\bigwedge_{i=1}^m\bigwedge_{j=i+1}^n h_j'dh_i=\prod_{i=1}^m t_{ii}^{n-i}(dT)(H_1'dH_1). \end{aligned}i=1∏mtiin−ii=1⋀mj=i+1⋀nhj′dhi=i=1∏mtiin−i(dT)(H1′dH1).
where (H1′dH1)=⋀i=1m⋀j=i+1n(H_1'dH_1)=\bigwedge_{i=1}^m\bigwedge_{j=i+1}^n(H1′dH1)=⋀i=1m⋀j=i+1n.
My comment: Why not (H′dH1)(H'dH_1)(H′dH1) ?
End of proof of (13).

For complex matrix ZZZ and unitary matrix H1:H2H_1:H_2H1:H2, considering the real part and imagine part,
(H2HdH1T)=(det⁡T)2n−2m⋀j=m+1n⋀i=1mhjHdhi.\begin{aligned} (H_2^H dH_1 T)=(\det T)^{2n-2m} \bigwedge_{j=m+1}^n \bigwedge_{i=1}^m h_j^Hdh_i. \end{aligned}(H2HdH1T)=(detT)2n−2mj=m+1⋀ni=1⋀mhjHdhi.

Now consider the upper matrix H1HdH1T+dTH_1^H dH_1 T+dTH1HdH1T+dT. First note that since H1HH1=ImH_1^H H_1=I_mH1HH1=Im, we have
H1HdH1+dH1H.H1=0\begin{aligned} H_1^H dH_1+dH_1^H.H_1=0 \end{aligned}H1HdH1+dH1H.H1=0
so that
H1HdH1=−dH1H.H1=−(H1HdH1)H,\begin{aligned} H_1^H dH_1=-dH_1^H.H_1=-(H_1^H dH_1)^H, \end{aligned}H1HdH1=−dH1H.H1=−(H1HdH1)H,
and hence H1HdH1H_1^H dH_1H1HdH1, is skew-symmetric:
H1HdH1=[h1Hdh1−h2Hdh1⋯−hmHdh1h2Hdh1h2Hdh2⋯−hmHdh2h3Hdh1h3Hdh2⋯−hmHdh3⋮⋮⋮hmHdh1hmHdh2hmHdhm]\begin{aligned} H_1^H dH_1=\begin{bmatrix} h_1^Hdh_1 & -h_2^Hdh_1 & \cdots & -h_m^Hdh_1\\ h_2^Hdh_1 & h_2^Hdh_2 & \cdots & -h_m^Hdh_2\\ h_3^Hdh_1 & h_3^Hdh_2 & \cdots & -h_m^Hdh_3\\ \vdots & \vdots & &\vdots\\ h_m^Hdh_1 & h_m^Hdh_2 & &h_m^Hdh_m \end{bmatrix} \end{aligned}H1HdH1=⎣⎢⎢⎢⎢⎢⎡h1Hdh1h2Hdh1h3Hdh1⋮hmHdh1−h2Hdh1h2Hdh2h3Hdh2⋮hmHdh2⋯⋯⋯−hmHdh1−hmHdh2−hmHdh3⋮hmHdhm⎦⎥⎥⎥⎥⎥⎤
where HiHdHi,(i=1,2,...,m)H_i^H dH_i, (i=1,2,...,m)HiHdHi,(i=1,2,...,m) are with imagine parts only.

Postmultiplying this by the upper-triangular matrix T gives the following matrix, where only the subdiagonal elements (and on the diaonal for complex matrix form)are given, and where, in addition, terms of the form hiHdhjh_i^Hdh_jhiHdhj are ignored if they have appeared already in a previous column:
H1HdH1T=[h1Hdh1t11∗⋯∗h2Hdh1t11h2Hdh2t22+∗⋯∗h3Hdh1t11h3Hdh2t22+∗⋯∗⋮⋮⋮⋮hmHdh1t11hmHdh2t22+∗⋯hmHdhmtmm+∗]\begin{aligned} H_1^H dH_1T=\begin{bmatrix} h_1^Hdh_1t_{11} & * & \cdots & *\\ h_2^Hdh_1t_{11} & h_2^Hdh_2t_{22}+* & \cdots & *\\ h_3^Hdh_1t_{11} & h_3^Hdh_2t_{22}+* & \cdots & *\\ \vdots & \vdots &\vdots&\vdots\\ h_m^Hdh_1t_{11} & h_m^Hdh_2t_{22}+* & \cdots & h_m^Hdh_mt_{mm}+* \end{bmatrix} \end{aligned}H1HdH1T=⎣⎢⎢⎢⎢⎢⎡h1Hdh1t11h2Hdh1t11h3Hdh1t11⋮hmHdh1t11∗h2Hdh2t22+∗h3Hdh2t22+∗⋮hmHdh2t22+∗⋯⋯⋯⋮⋯∗∗∗⋮hmHdhmtmm+∗⎦⎥⎥⎥⎥⎥⎤

Column by column, the exterior product of the subdiagonal elements (under the diagonal)of H1′dH1T+dTH_1' dH_1 T+dTH1′dH1T+dT is (remember that dTdTdT is upper-triangular)

t112(m−1)+1t222(m−2)+1⋯tm,m⋀i=1m⋀j=i+1mhj′dhi.\begin{aligned} t_{11}^{2(m-1)+1}t_{22}^{2(m-2)+1}\cdots t_{m,m}\bigwedge_{i=1}^m\bigwedge_{j=i+1}^m h_j'dh_i. \end{aligned}t112(m−1)+1t222(m−2)+1⋯tm,mi=1⋀mj=i+1⋀mhj′dhi.
Each element under diagonal contribute 2 tiit_{ii}tii, and the element on diagonal contribute 1 tiit_{ii}tii.
(H1HdH1T+dT)=t112m−1t222m−3⋯tm,m⋀i=1m⋀j=imhjHdhi(dT).\begin{aligned} (H_1^H dH_1T+dT)=t_{11}^{2m-1}t_{22}^{2m-3}\cdots t_{m,m}\bigwedge_{i=1}^m\bigwedge_{j=i}^m h_j^Hdh_i(dT). \end{aligned}(H1HdH1T+dT)=t112m−1t222m−3⋯tm,mi=1⋀mj=i⋀mhjHdhi(dT).

The exterior product of the elements of H2HdH1TH_2^H dH_1 TH2HdH1T
and the subdiagonal elements (and imagine part on diagonal , considering dtiidt_{ii}dtii are real) of H1HdH1T+dTH_1^H dH_1 T+dTH1HdH1T+dT is
∏i=1mtiin−i⋀i=1m⋀j=inhjHdhi=∏i=1mtiin−i(H1HdH1).\begin{aligned} \prod_{i=1}^m t_{ii}^{n-i}\bigwedge_{i=1}^m\bigwedge_{j=i}^n h_j^Hdh_i=\prod_{i=1}^m t_{ii}^{n-i}(H_1^HdH_1). \end{aligned}i=1∏mtiin−ii=1⋀mj=i⋀nhjHdhi=i=1∏mtiin−i(H1HdH1).
Here (H1HdH1)=⋀i=1m⋀j=inhjHdhi(H_1^HdH_1)=\bigwedge_{i=1}^m\bigwedge_{j=i}^n h_j^Hdh_i(H1HdH1)=⋀i=1m⋀j=inhjHdhi.

The exterior product of the elements of H1′dH1T+dTH_1' dH_1 T+dTH1′dH1T+dT on diagnal (real part) and above the diagonal is
⋀i≤jm+terms involving dH1.\begin{aligned} \bigwedge_{i\le j}^m + \textrm {terms involving }{d}H_1. \end{aligned}i≤j⋀m+terms involving dH1.
The terms of dH1{d}H_1dH1 do not contribute to the total exterior product.Hence the exterior product of the elements of the right side is
∏i=1mtii2n−2i+1⋀i=1m⋀j=inhjHdhi=∏i=1mtii2n−2i+1(dT)(H1HdH1).\begin{aligned} \prod_{i=1}^m t_{ii}^{2n-2i+1}\bigwedge_{i=1}^m\bigwedge_{j=i}^n h_j^Hdh_i=\prod_{i=1}^m t_{ii}^{2n-2i+1}(dT)(H_1^HdH_1). \end{aligned}i=1∏mtii2n−2i+1i=1⋀mj=i⋀nhjHdhi=i=1∏mtii2n−2i+1(dT)(H1HdH1).

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