Theorem 12.6.12.B01FA2265C81 The metric space $(S_ W,d_ W)$ is complete and proper, and a sequence in $S_ W$ is Cauchy exactly when its coefficient matrices converge to a positive matrix in $S_ W$. For $A,B\in S_ W$,
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\begin{align} d_ W(A,B)& \geq \frac1{\sqrt n} \| \log (A^{-1/2}BA^{-1/2})\| _{\mathrm F},\\ d_ W(A,B)& \leq \frac{\| A-B\| _{\mathrm F}}{\sqrt n\min \{ \lambda _{\min }(A),\lambda _{\min }(B)\} }. \tag{12.6.14} \end{align}
The normalized minimizing metrics of (12.6.6) satisfy, pointwise on $X$,
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\begin{equation} \mathrm e^{-\sqrt n d_ W(A,B)}\gamma _ A \leq \gamma _ B\leq \mathrm e^{\sqrt n d_ W(A,B)}\gamma _ A. \end{equation}
In particular, the completion of $S_ W$ adds no point, and no point of the boundary of the coefficient cone is a limit of a Cauchy sequence.
If $A_ j>0$ in $W$ converges to a nonzero singular semipositive $A_0$, then $\widehat A_ j=A_ j/\lambda (A_ j)$ lies in $S_ W$ and, for fixed $A_\star \in S_ W$,
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\begin{equation} d_ W(A_\star ,\widehat A_ j) \geq \frac{|\log \det A_ j|}{n\sqrt n}+O(1) \longrightarrow \infty . \end{equation}
By contrast, a sequence of coefficient matrices tending to zero can normalize to a point of $S_ W$.
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
\[ U=B-A,\qquad C(t)=A+tU,\qquad N(t)=\frac{C(t)}{\lambda (C(t))}. \]
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
\[ \begin{gathered} g_ N(N',N')=\frac1n\operatorname {tr}(K^2) -\frac1{n^2}(\operatorname {tr}K)^2 \leq \frac{\| U\| _{\mathrm F}^2}{nm^2},\\ m=\min \{ \lambda _{\min }(A),\lambda _{\min }(B)\} . \end{gathered} \]
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
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\begin{equation} \mathrm e^{-\sqrt n d}A\leq B\leq \mathrm e^{\sqrt n d}A. \end{equation}
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
\[ \| A_ j-A_ k\| _{\mathrm{op}} \leq M\bigl (\mathrm e^{\sqrt n d_ W(A_ j,A_ k)}-1\bigr ), \]
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
\[ \begin{gathered} \mathrm e^{-\sqrt nR}A_\star \leq A\leq \mathrm e^{\sqrt nR}A_\star ,\\ A\in W,\qquad H\det A=1. \end{gathered} \]
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
\[ \begin{aligned} d_ W(A_\star ,\widehat A_ j) & \geq \frac1{\sqrt n}\log \frac{\lambda _{\max }(A_ j)}{\lambda (A_ j)\lambda _{\max }(A_\star )}\\ & =\frac{|\log \det A_ j|}{n\sqrt n}+O(1). \end{aligned} \]
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.
$\square$
Pome
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