12.6.4 Invariant backgrounds on compact Lie-group quotients8D8E5D56DAEF

Proof. Invariant averages and balancedness. Let $Z_1,\ldots ,Z_ n$ be the invariant holomorphic vector fields dual to $\alpha _1,\ldots ,\alpha _ n$, and set $\Upsilon =\alpha _1\wedge \cdots \wedge \alpha _ n$. This holomorphic top-degree form and its conjugate are closed. For an invariant holomorphic vector field $Z$, evaluation on the invariant frame gives

The second identity also follows from Cartan's formula. Since $h_0^ n$ is a positive constant multiple of the appropriately oriented form $\Upsilon \wedge \bar\Upsilon $, Stokes' theorem on compact $X$ gives

Thus the fields $Z_ j$ and their conjugates have zero divergence with respect to $\mu $. In particular,

Cartan's formula also gives $\mathrm d(\iota _{Z_ j}\Upsilon )=0$. These contractions span the invariant $(n-1,0)$-forms. The form $\theta _ A^{n-1}$ is a constant linear combination of their wedges with conjugates, so $\mathrm d\theta _ A^{n-1}=0$.

The holomorphic fields commute with the antiholomorphic fields, and $\mathrm d\alpha _ j$ has type $(2,0)$. The coefficient matrix of $\mathrm{dd}^{\mathrm{c}}u$ is therefore $(Z_ j\bar Z_ k u)$. For smooth $u$ its entries have $\mu $-mean zero by (12.6.8), so the matrix of $\mu $-means of the coefficients of $\theta _ A+\mathrm{dd}^{\mathrm{c}}u$ is $A$. If $\theta _ A+\mathrm{dd}^{\mathrm{c}}u>0$, averaging the positive coefficient matrix gives $A>0$. The converse follows by taking $u=0$. The same average shows that $V_ W$ contains no nonzero smooth $\mathrm{dd}^{\mathrm{c}}$ direction.

For a distribution $u$, the adjoints of $Z_ j$ and $\bar Z_ k$ with respect to $\mu $ annihilate constants, so the mean of each $Z_ j\bar Z_ k u$ is still zero. The coefficients of a positive current in this global frame form a positive matrix of measures, whose matrix of total masses is semipositive; for $\theta _ A+\mathrm{dd}^{\mathrm{c}}u\geq 0$ this gives $A\geq 0$. Conversely, $\theta _ A$ itself is a positive current when $A\geq 0$. If $A=0$, the positive trace measure of $\mathrm{dd}^{\mathrm{c}}u$ has total mass zero, and polarization then makes the entire current zero. Finally, for a fixed positive $A_+\in W$, the matrices $A+\varepsilon A_+$ with $\varepsilon \downarrow 0$ approximate every semipositive $A\in W$ from the positive cone. This proves the closure assertion.

The quotient and its equality case. Put $Q=H\det A$, so that $\theta _ A^ n=Q\mu $. For every smooth positive Hermitian form $\gamma $, the matrix arithmetic–geometric mean inequality gives

Integration yields $\int _ X\theta _ A\wedge \gamma ^{n-1}\geq Q^{1/n}D_\mu (\gamma )$, which bounds every Gauduchon quotient in definition 12.2.1 from below. The balanced, hence Gauduchon, metric $\gamma =\theta _ A$ has $D_\mu (\theta _ A)=Q^{(n-1)/n}$ and attains equality. This proves the formula for $\lambda _\mu (\theta _ A)$.

For a minimizing Gauduchon metric $\gamma $, the continuous nonnegative difference of the two sides of the pointwise inequality has integral zero. Pointwise equality then forces $\gamma =f\theta _ A$ for a smooth positive function $f$. Set $q=f^{n-1}$ and $T=\theta _ A^{n-1}$. Since $\mathrm dT=0$, the Gauduchon equation becomes $\mathrm{dd}^{\mathrm{c}}q\wedge T=0$, and

The last integrand is nonnegative and vanishes exactly where $\mathrm dq=0$. Thus $f$ is constant. As $n\geq 2$ and

normalization determines $c=\lambda _\mu (\theta _ A)^{-1}$. This proves the unique minimizer and its volume identity.

Differentiating $-\log \lambda _\mu (\theta _ A)=-(\log H+\log \det A)/n$ twice gives (12.6.7). The corresponding squared norm of $U$ is $\| A^{-1/2}UA^{-1/2}\| _{\mathrm F}^2/n$, which is positive for $U\ne 0$; here $\| B\| _{\mathrm F}^2=\operatorname {tr}(B^*B)$ denotes the squared Frobenius norm.

Finally,

For invertible $A$, the vanishing of this form forces every $\mathrm d\alpha _ j=0$, by independence of the conjugate coframe, and the Lie algebra would then be abelian. Since adding $\mathrm{dd}^{\mathrm{c}}u$ does not change $\mathrm d\theta _ A$, this proves the last assertion. $\square$

Proof. By the trace formula, the congruence $M\mapsto P^*MP$ by a fixed invertible matrix $P$ is an isometry. Taking $P=A^{-1/2}$, we may assume that the endpoints are $I$ and $C=A^{-1/2}BA^{-1/2}>0$. The curve $\exp (t\log C)$, $0\leq t\leq 1$, has constant speed $\| \log C\| _{\mathrm F}/\sqrt n$, which gives the upper bound for the distance.

For the reverse inequality, diagonalization gives the convergent resolvent formula for the matrix logarithm and its derivative

On a neighborhood of $M$ where the spectrum is bounded below by a positive constant, differentiation under the integral sign is justified by an integrable bound. In a unitary basis diagonalizing $M$, with eigenvalues $d_ i>0$, the derivative has entries $c_{ij}U_{ij}$, where

Writing $d_ i/d_ j=\mathrm e^{2s}$ gives $\sqrt{d_ id_ j}|c_{ij}|=|s/\sinh s|\leq 1$, including the limiting value at $s=0$. Hence

Applying this along a piecewise $C^1$ path of positive matrices from $I$ to $C$ and integrating, we bound its length below by $\| \log C\| _{\mathrm F}/\sqrt n$. This proves the distance formula. In terms of the original endpoints, the length-minimizing curve is

Its determinant is $(\det A)^{1-t}(\det B)^ t$, so for endpoints with equal determinant it stays in the fixed-determinant hypersurface. $\square$

Proof. Paths in $W$. Every path in $S_ W$ is a path in the hypersurface of all positive matrices with $H\det A=1$, so lemma 12.6.9 gives (12.6.13). To obtain an upper bound by paths inside $W$, put

Since the positive cone is convex and $W$ is linear, $N(t)$ stays in $S_ W$ and joins $A$ to $B$. With $K=C^{-1/2}UC^{-1/2}$, differentiation gives

Indeed $C(t)\geq mI$ and $(\log \lambda (C))'=\operatorname {tr}(C^{-1}U)/n$. Integrating proves (12.6.14), and shows in particular that $S_ W$ is connected.

Put $d=d_ W(A,B)$. Every eigenvalue of $A^{-1/2}BA^{-1/2}$ has a logarithm of absolute value at most $\| \log (A^{-1/2}BA^{-1/2})\| _{\mathrm F}\leq \sqrt n\, d$, by (12.6.13). Hence

Since $\gamma _ A=\theta _ A$ on $S_ W$, this proves (12.6.15).

Cauchy sequences and bounded balls. For a Cauchy sequence $(A_ j)$ in $(S_ W,d_ W)$, comparison with one fixed element of the sequence, using (12.6.17), gives $mI\leq A_ j\leq MI$ with fixed positive $m,M$. Comparing two elements then gives

where $\| \cdot \| _{\mathrm{op}}$ is the operator norm. The coefficient matrices therefore form a Cauchy sequence. Their limit lies in the closed linear space $W$, is positive, and satisfies $H\det A=1$, and the upper bound (12.6.14) proves convergence to that point in $d_ W$. Conversely, coefficient convergence to a positive point gives convergence in $d_ W$ by the same bound. This proves completeness and the Cauchy criterion. The minimizing forms then converge in $C^\infty $, because their coefficients multiply a fixed smooth coframe.

By these estimates, $d_ W$ induces the topology of coefficient convergence. A closed ball of radius $R$ around $A_\star $ is contained in the compact matrix interval

The ball is closed in that interval, hence compact. This proves properness. In particular, distances from a fixed reference point are bounded exactly when the eigenvalues relative to that point are bounded above and away from zero.

The coefficient boundary. Write $\lambda _{\max }$ for the largest eigenvalue. If $A_ j\to A_0$ with $A_0\geq 0$ singular and nonzero, then $\lambda (A_ j)\to 0$ while $\lambda _{\max }(A_ j)$ has a positive limit, so the largest eigenvalue of $\widehat A_ j$ tends to infinity. Applying (12.6.17) to $A_\star ,\widehat A_ j$ gives

The error term is bounded because $A_0\ne 0$. This proves (12.6.16). For comparison, $A_ j=t_ jA_\star $ with $t_ j\downarrow 0$ normalizes to $A_\star $ for every $j$; its coefficient limit, the zero matrix, gives no new point of the completion. $\square$

Proof. The bilinear cross term makes the multiplication associative, with identity zero and inverse $(-z_1,-z_2,-z_3+z_1z_2)$. These operations preserve the Gaussian lattice. Its action by left multiplication is free and properly discontinuous: for a compact set $K\subset G$, elements $\gamma $ with $\gamma K\cap K\ne \varnothing $ belong to the compact set $KK^{-1}$, containing only finitely many lattice elements.

Choose Gaussian integers $\gamma _1,\gamma _2$ that move $z_1,z_2$ into the half-open unit square, and then $\gamma _3$ that moves $z_3+\gamma _1z_2$ into it. This gives a representative in the real six-dimensional unit cube. If two points of the half-open cube are equivalent, then $\gamma _1=\gamma _2=0$ and then $\gamma _3=0$. The cube is therefore a measurable fundamental domain, and $X$ is compact, being the image of its closure.

This coframe is left-invariant, descends to $X$, and has

The density of $h_0^3$ is $6/\pi ^3$ times real Lebesgue volume on the fundamental cube, giving the stated $H$. The coefficient of $\alpha _1\wedge \alpha _2\wedge \bar\alpha _3$ in $\partial \theta _ A$ is $-\mathrm iA_{33}/(2\pi )$, so positivity makes every such background nonclosed.

To see that $X$ admits no Kähler metric at all, suppose that a positive closed $(1,1)$-form $\kappa $ existed, and average its coefficient functions in this coframe against $\mu $. Their mean is a positive matrix $A$. In the coefficients of $\partial \kappa $, derivative terms have mean zero by (12.6.8); the remaining means are exactly the coefficients of $\partial \theta _ A$. Closedness would force them to vanish, contradicting $A_{33}>0$. Thus $X$ is non-Kähler.

The assertions about the volume, the minimizers and completeness follow from proposition 12.6.5, theorem 12.6.12. As $W$ is the full matrix space, the equal-determinant exponential path stays in $S$, and lemma 12.6.9 gives (12.6.19). For $A_\star =aI$, $a=(\pi ^3/6)^{1/3}$, this becomes

The three logarithms sum to zero.

For a concrete degeneration, take $A_\varepsilon =\operatorname {diag}(1,1,\varepsilon )$, $0<\varepsilon \leq 1$. As $\varepsilon \to 0$, the coefficients converge to the nonzero smooth semipositive form $\theta _{\operatorname {diag}(1,1,0)}$. The normalized matrices and their distances, on the other hand, are

Thus, after normalization, the boundary form $\theta _{\operatorname {diag}(1,1,0)}$ lies at infinite distance. $\square$