数理解析研究所講究録
第 1632 巻 2009 年 28-37

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正作用素の微分幾何と計量

Differential geometries and their metrics
on the positive operators
大阪教育大学・教養学科・情報科学
藤井 淳一 (Jun Ichi Fujii )Departments of Arts and Sciences (Information Science)

Osaka Kyoiku University

1

Introduction

To discuss geometries with metrics induced by unitarily invariant norms, let $A$
be a unital-algebra on a finite dimensional Hilbert spaoe $H$ . The manifold in, the positive invertible elements. In $1990’ s$ , Corach-Porta-Lecht
discourse is[7, 8, 9] discussed Finsler geometry on it, which we call the $CPR$ geometry: Thetangent space (and bundle) is the selfadjoint elements. The invertible elements$A$
in defines the principal fiber (frame) bundle: For a fixed, the projection$\pi_{A}(G)=GAG^{*}$ , the structure group $\mathcal{U}_{A}=\{V\in \mathcal{G}|VAV^{*}=A\}=A^{1/2}\mathcal{U}A^{-1/2}$
with the action $L_{V}A=VAV^{*}$ , which shows
is homogeneous and hence canbe assumed the identity element . In this caseis nothing but the unitary group.
itself is considered as the tangent space of and it has the connection inducedby the horizontal space. Then, the parallel displacement of a tangent
vector $X$ along $\gamma hom0$ to is$C^{*}$$\mathcal{A}^{+}$$\mathcal{A}^{h}$$A\in \mathcal{A}$$\mathcal{G}$$A$$\mathcal{A}^{+}$$I$$\mathcal{U}$$\mathcal{U}_{I}$$\mathcal{A}$$\mathcal{G}$$H_{G}=G\mathcal{A}^{h}$$t$$P_{t}X=\Gamma(t)\Gamma(0)^{-1}X(\Gamma(0)^{*})^{-1}\Gamma(t)^{*}$

So the covariant denvative$D_{t}$.

of a tangent field $X(t)$ along the curve$\gamma(t)$in$\mathcal{A}^{+}$

isgiven by$D_{t}X= \dot{X}-\frac{1}{2}(\dot{\gamma}\gamma^{-1}X+X\gamma^{-1}\dot{\gamma})$.

Then the geodesic equation
implies that the geodesic fromtois the path of geometric Kubo-Ando means [14]:$O=D_{t}\dot{\gamma}=\ddot{\gamma}-\dot{\gamma}\gamma^{-1}\dot{\gamma}$

$A$

$B$

$\gamma(t)=A\neq tB=A^{\underline{\iota_{\nabla}}}I(A^{-1}zBA^{-f})^{t}A1\xi$

Moreover the above manifold$\mathcal{A}^{+}$.

is the Finsler space with a Finsler metric$L(X;A)=\Vert X\Vert_{A}=\Vert A^{-1/2}XA^{-1/2}\Vert$at each point
and the distance between two points defined as the shortest
length of a path is so-called Thompson metric $d(A, B)=\Vert\log A^{-1/2}BA^{-1/2}\Vert$ .
The CPR geometry does not always determine a unique Finsler metric. In fact,
we show each unitarily invariant norm 1 also gives a Finsler metric for the CPR
geometry:$A\in \mathcal{A}$$\Vert|$29
Theorem 1. For a unitarily invariant norm 1$\Vert|$on A. a function$L_{\Vert 1}\Vert|(X;arrow\prime i)=\Vert|X\Vert|_{A}=\Vert|A^{-1/2}XA^{-1/2}\Vert|$

determines a Finsler metric on

Proof. Since$U_{t}=\gamma(t)^{-1/2}\Gamma(t)$$\mathcal{A}^{+}$

for the $CPR$ geometry.

defines a unitary for each by$t$$\gamma=$rr“, we show theFinsler condition by$\Vert|P_{t}X\Vert|_{\gamma(t)}=\Vert|U_{t}U_{0}^{*}\gamma(0)^{-1/2}X\gamma(0)^{-1/2}U_{0}U_{t}^{*}\Vert|=\Vert|\gamma(0)^{-1/2}X\gamma(0)^{-1/2}\Vert|=\Vert|X\Vert|_{\gamma(0)}$.

口

A Finsler metrichomogeneous like $L(X;A)$ :$L_{\Vert 1}\Vert|(X;A)$. which is called a unitarily invari ant Finsler one, is

Theorem 2. For any invertible operator ,$Y$$L_{\Vert 1}\Vert|(Y^{*}XY;Y^{*}AY)=L_{1\Vert}\Vert|(X;A)$

Proof. Since$\Vert|Z\Vert|=\Vert||Z|\Vert|=\Vert|\sqrt{Z^{*}Z}\Vert|=\Vert|\sqrt{ZZ^{*}}\Vert|$.

, we have$L_{\Vert|\Vert|}(Y^{*}XY;Y^{*}AY)=\Vert|(Y^{*}AY)^{-1/2}Y^{*}XY(Y^{*}AY)^{-1/2}\Vert|$$==\Vert|_{\sqrt{(YAY)/YXAXY(YAY)/}\Vert 1}\sqrt{(YAY)/YXY(YAY)YXY(YAY)/}\Vert|$$==\Vert|_{\sqrt{A/XAXA}}^{\sqrt{A/XY(YAY)YXA/}\Vert 1}\Vert|=\Vert|A^{-1/2}XA^{-1/2}\Vert|=L_{\Vert|\Vert|}(X;A$口2Metric space of Thompson’s typeThe geodesic $A\# tB$ is one of the shortest paths with respect to this metric: The
lengthof pathis defined by$\ell(\gamma)$$\gamma(t)$$l( \gamma)\equiv\int_{0}^{1}L(\gamma’(t);\gamma(t))dt=\int_{0}^{1}\Vert\gamma(t)^{-1/2}\gamma’(t)\gamma(t)^{-1/2}\Vert dt$

If$\gamma(t)$is a path from$A$

to $B$ , then$d(A, B) \equiv\inf_{\gamma}l(\gamma)=\ell(A\neq tB)=\Vert\log(A^{-1/2}BA^{-1/2})\Vert$$= \log(\max\{||A^{-1/2}BA^{-1/2}\Vert, \Vert B^{-1/2}AB^{-1/2}\Vert\})$$= \log(\max\{r(A^{-1}B), r(B^{-1}A)\})$

.

.

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The homogeneity of$\mathcal{A}^{+}$

implies$d(-4, B)=d(Y^{*}AX.X^{*}BX)=d(I.A^{-1/2}BA^{-1/2})$

for invertible $X$ . The metric makesa complete metric space and it is called
the Thompson (part) one [19].
Also. Corach et.al. [8, 3] showed the convexity for the metric: For geodesics
and , the followings are equivalent:$d$$\mathcal{A}^{+}$$\gamma$$\delta$: convex:(i)$F(t)=d(\gamma(t), \delta(t))=\log||\gamma(t)^{-1/2}\delta(t)\gamma(t)^{-1/2}||$(ii)$d(\gamma(t), \delta(t))\leqq(1-t)d(\gamma(O), \delta(0))+td(\gamma(1), \delta(1))$.

The above equivalence is guaranteed by the interpolationality for the path $A\# tB$ .
This convexity suggests that the curvature ofis negative. In Riemannian geometry, the above convexity implies exactly the negativity of the curvature. But, inFinsler geometry, the notion of it has not been completely established yet.
Now we show the above properties also hold for the metric$\mathcal{A}^{+}$$d_{\Vert 1}[(A, B)= \inf_{\gamma\in P(AB)},\int_{0}^{1}L_{|\Vert\Vert|}(\gamma^{l}(t);\gamma(t))dt$

for a unitarily invariant norm 1
where $P(A, B)$ denotes the (differentiable) paths
from to $B$ . To see the path $A\# tB$ is one of the shortest paths directly, recall thefollowing ‘logarithmic-geometric mean inequality’ due to Hiai-Kosaki [13]:$\Vert|$$A$

Hiai-Kosaki inequality.

1

Theorem 2 implies the homogeneity of$\int_{0}^{1}H^{t}XK^{1-t}dt\Vert|\geqq\Vert|H^{1/2}XK^{1/2}\Vert|$.

$d_{1\Vert}\Vert^{;}$

Lemma 3. $d_{N}|\Vert(A, B)=d_{\Uparrow 1}\Vert|(X^{*}AX, X^{*}BX)$ holds for inventble $X$ .

Proof. Note that $X^{*}P(A, B)X=P(X^{*}AX, X^{*}BX)$ holds for all invertible$L_{\Downarrow 1}\Vert|((X^{*}\gamma’(t)X;X^{*}\gamma(t)X)=L_{1\Vert}\Vert|(\gamma’(t);\gamma(t))$$X$. Since, we have$d_{Y1}[(X^{*}AX, X^{*}BX)= \gamma\in P(X^{*}AXXBX)\inf_{1}.\int_{0}^{1}L_{|\Vert|1}(\gamma’(t);\gamma(t))dt$$= \inf_{\gamma\in P(AB)},\int_{0}^{1}L_{N1}\#(X^{*}\gamma’(t)X;X^{*}\gamma(t)X)dt$$= \inf_{\gamma\in P(AB)},\int_{0}^{1}L_{[}\#(\gamma’(t);\gamma(t))dt=d_{N1[}(A, B)$

by Theorem 2.

The following formula is parallel to Bhatia’s one [6, (6.42)].

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Lemma 4. Let $H(t)=\log\gamma(t)$ for$\gamma\in P(A_{\dot{\text{佰}}}B)$

Then

.

.

$\frac{d}{(tt}\gamma(t)=\frac{d}{dt}e^{H(t)}=\int_{0}^{1}e^{uH(t)}H’(t)e^{(1-u)H(t)}du$

Proof. Since$- \frac{d}{du}e^{uH(t)}e^{(1-u)H(t+\Xi)}=e^{uH(t)}(H(t+\epsilon)-H(t))e^{(1-u)H(t+\epsilon)}$,we have$\int_{0}^{1}e^{uH(t)}(H(t+\epsilon)-H(t))e^{(1-u)H(t+e)}du=-[e^{uH(t)}e^{(1-u)H(t+\epsilon)}]_{0}^{1}=e^{H(t+\epsilon)}-e^{H(t)}$The require formula is obtained by multiplying

and$\frac{1}{\epsilon}$$\epsilonarrow 0$

.

.

ロ

Now we show that it is a metric of Thompson type, which is mentioned also in [5],and moreover that the geodesic $A\# tB$ is the shortest path in almost all cases. To
see this, recall that the norm 1 is strictly convex if$\Vert|$for all $t\in(O, 1)$ and all distinct unit vectors$\Vert|(1-t)x+ty\Vert|<1$

$x$

and .
$y$

Then the strict triangle inequality holds (cf. [17]):$\Vert|x+y\Vert|<\Vert|x\Vert|+\Vert|y\Vert|$

unless one of the vectors is a nonnnegative multiple of the other.
Note that we identify
with
if is increasing function with $f(O)=1$
and $f(1)=1$ since they are the same as sets; $\{\gamma(f(t))|t\in[0,1]\}=\{\gamma(t)|t\in[0,1]\}$and give the same length: $\ell(\gamma(f(t)))=\ell(\gamma(t))$ .
$\gamma(f(t))$

Theorem 5. A function$d_{1\Vert}\Vert|$$\gamma(t)$is a metric on$d_{1\Vert} \Vert|(A, B)=\int_{0}^{1}L_{1\Vert}|\Vert(\frac{d}{dt}A\# tB,$

Moreover, if the norm
path.

Proof. First we see$\Vert|\Vert|$$f$$\mathcal{A}^{+}$

and$A\# tB)dt=\Vert|\log A^{-1/2}BA^{-1/2}\Vert|$.

is strictly convex, then the geodesic $A\# tB$ is the shortest$d_{1\Vert}\Vert|(A, B)\geqq\Vert|\log A^{-1/2}BA^{-1/2}\Vert|$. In fact, by Lemma 4 and theHiai-Kosaki inequality,$\Vert|\gamma(t)^{-1/2}\gamma’(t)\gamma(t)^{-1/2}\Vert|=\Vert|e^{-H(t)/2}(\int_{0}^{1}e^{uH(t)}H’(t)e^{(1-u)H(t)}du)e^{-H(t)/2}\Vert|$$= \Vert|(\int_{0}^{1}(e^{H(t)})^{u}e^{-H(t)/2}H’(t)e^{-H(t)/2}(e^{H(t)})^{1-u}du)\Vert|$$\geqq\Vert|(e^{H(t)})^{1/2}e^{-H(t)/2}H’(t)e^{-H(t)/2}(e^{H(t)})^{1/2}\Vert|=\Vert|H’(t)\Vert|$

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holds and hence thc required inequality is obtaiiied by$\ell_{\Vert 1}\Vert|(\gamma)=\int_{0}^{1}\Vert|\gamma^{-1/2}(t)\gamma’(t)\gamma^{-1/2}(t)\Vert|dt$.

$\geqq\int_{0}^{1}\Vert|H’(t)\Vert|dt\geqq\Vert|\int_{0}^{1}H’(t)dt\Vert|=\Vert|\log B-\log A\Vert|$

Let $C=A^{-1/2}BA^{-1/2}$ and $\Gamma(t)=C^{t}$ . Then$\ell_{\Vert 1}||(\Gamma)\equiv\int_{0}^{1}\Vert|C^{-t/2}(C^{t}\log C)C^{-t/2}\Vert|dt=\Vert|\log C\Vert|=\Vert|\log C-\log I\Vert|$

and hence attains the shortest length and $d(I, C)=\Vert|\log C\Vert|$ . By Lemma 3, wehave the geodesic $\gamma(t)=A^{1/2}\Gamma(t)A^{1/2}=A^{1/2}C^{t}A^{1/2}=A\# tB$ . Moreover,$\Gamma$$d_{1\Vert}\Vert|(A, B)=d_{\Vert 1}\Vert|(I, C)=\Vert|\log C\Vert|=\Vert|\log A^{-1/2}BA^{-1/2}\Vert|$$=\Vert|\log B^{-1/2}AB^{-1/2}\Vert|=d_{1\Vert}|\Vert(B, A)$.

Thus the symmetry of $d_{M}\Vert|(A, B)$ holds. It is clear that $d_{\Vert 1}|\Vert(A, B)=0$ if and only if$A=B$ and that the triangle inequality forholds, which shows
is a metric.
Next suppose the norm is strictly convex and attains the shortest. By homegeneity, we may assume $\gamma\in P(I, B)$ . Then, for $H(t)=\log\gamma(t)$ , it must satisfy$d_{\Vert 1}| \int$$d_{\Vert 1}\Vert|$$\gamma$$\int_{0}^{1}\Vert|H’(t)\Vert|dt=\Vert|\int_{0}^{1}H’(t)dt\Vert|=\Vert|\log B\Vert|$.

Thereby we have $H’(t)$ is a nonnegative scalar multiple of $\log B$ for each . In fact,
we use the broken line approxmation to obtain the length of $H(t)$ :$t$.

$\int_{0}^{1}\Vert|H’(t)\Vert|dt=\lim_{|\Delta|arrow 0}\sum_{t_{n}\in\Delta}\Vert|H(t_{n+1})-H(t_{n})\Vert|$

Take the following monotone increasing sequence converging to$\sum_{k=1}^{2^{n}}\Vert|H(\frac{k}{2^{n}})-H(\frac{k-1}{2^{n}})\Vert|$$\uparrow$$\int_{0}^{1}\Vert H’(t)\Vert dt$$\int_{0}^{1}\Vert|H’(t)\Vert|dt$:.

Then all the triangle inequalities are equal: e.g,,$\Vert|H(\frac{k}{2^{n}})-H(\frac{k-1}{2^{n}})\Vert|$$\leqq\Vert|H(\frac{2k}{2^{n+1}})-H(\frac{2k-1}{2^{n+1}})\Vert|+\Vert|H(\frac{2k-1}{2^{n+1}})-H(\frac{2(k-1)}{2^{n+1}})\Vert|$.

For $n=0$ and $k=1$ , we obtain $H(1)-H(1/2)=s(H(1/2)-H(O))=sH(1/2)$ ,that is, $H(1)=(s+1)H(1/2)$ . Putting $f(1/2)=1/(1+s)$ , we have $H(1/2)=$

33$f(1/2)H(1)=f(1/2)\log B$ . Since $H$ is continuous, we can define $f$ for all $t\in[0.1]$

and$H(t)=f(t)\log B=\log B^{f(t)}$Then $f(1)=0$ and $f(1)=1$ and moreoverso that is the same as the geodesic .

$f$

$ie.$

.

$\gamma(t)=B^{f(t)}$.

must be monotone, hence increasing,$B^{t}$$\gamma$$\square$Remark . If the norm is the operator one or the trace one, then it is not strictly
convex and indeed the shortest path is not uniquely determined. In fact, take apath $I\nabla_{t}B=(1-t)I+tB\in P(I,\cdot B)$ for $B\geq I$ . Then, considering$1^{\backslash }$$L_{\Vert 1}\Vert|$$(\dot{\gamma};\gamma)=\Vert|(B-I)(I+t(B-I))\Vert|$

and a function $F(x)=(x/(1+tx))$ is monotone increasing, we obtain $l(I\nabla_{t}B)=$in both cases.
$\Vert|\log B\Vert|$

Remark 2. In the above metric, it is easy to see thatis also complete. In$\Vert|P\Vert|=1$
particular, if a unitarily invariant norm is normalized;
for projections ofrank one, thenholds and hence the convergence is reduced to theoriginal Thompson metric.
$\mathcal{A}^{+}$$d_{\Vert 1}\Vert|$$\Vert X\Vert\leqq\Vert|X\Vert|$

Next we see the following equivalence:
Theorem 6. For geodesics
(i)
(ii)
(iii)

and , the followings hold and they are equivalent:$\delta$$\gamma$$F(t)=d_{1\Vert}\Vert|(\gamma(t), \delta(t))=\Vert|\log\gamma(t)^{-1/2}\delta(t)\gamma(t)^{-1/2}\Vert|$$d_{1\Vert}\Vert|(A^{t}, B^{t})\leqq td_{1\Vert}\Vert|(A, B)$

is convex.

.

$d_{1\Vert}\Vert|(\gamma(t), \delta(t))\leqq(1-t)d_{1\Vert}\Vert|(\gamma(0), \delta(0))+td_{1\Vert}\Vert|(\gamma(1), \delta(1))$.

Proof. It suffices to show the case of matrices. Then Araki [2] showed$\prod_{j=1}^{k}\lambda_{j}(B^{-t/2}A^{t}B^{-t/2})\leqq\prod_{j=1}^{k}\lambda_{j}^{t}(B^{-1/2}AB^{-1/2})$

for $0\leqq t\leqq 1$ and $1\leqq k\leqq n$ whereis the j-th eigenvalue (singular value in thiscase) under the decreasing order. Thus we have the weak-majorization$\lambda_{j}$(1)$\log B^{-t/2}A^{t}B^{-t/2}\prec wt\log B^{-1/2}AB^{-1/2}$

On the other hand, Araki’s inequality also holds for$A^{-1}$.

and$B^{-1}$;$\prod_{j=1}^{k}\lambda_{j}((B^{-t/2}A^{t}B^{-t/2})^{-1})=\prod_{j=1}^{k}\lambda_{j}(B^{t/2}A^{-t}B^{t/2})$$\leqq\prod_{j=1}^{k}\lambda_{j}^{t}(B^{1/2}A^{-1}B^{1/2})=\prod_{j=1}^{k}\lambda_{j}^{l}((B^{-1/2}AB^{-1/2})^{-1})$,

34
so that(2)$-\log B^{-t/2}A^{t}B^{-t/2}\prec-t\log B^{-1/2}AB^{-1/2}w$.

Combining (1) and (2), we have the majorization$\log B^{-t/2}A^{t}B^{-t/2}\prec t\log B^{-1/2}AB^{-1/2}$

and consequently we have (ii) by the convexity of $f(x)=|x|$ :$d_{1\Vert}\Vert|(A^{t}, B^{t})=\Vert|\log B^{-t/2}A^{t}B^{-t/2}\Vert|\leqq t\Vert|\log B^{-1/2}AB^{-1/2}\Vert|=td_{\Vert 1}\Vert|(A, B)$.

Then, the triangle inequality and the homogeneity (Lemma 3) and (ii) show (iti):

For $\gamma(t)=Am_{t}B,$ $\delta(t)=Cm_{t}Dand\zeta(t)=Cm_{t}B$ ,$d_{\Vert 1}\Downarrow(\gamma(t), \delta(t))\leqq d_{\Vert 1}\Vert|(\gamma(t), \zeta(t))+d_{\Vert 1}||(\zeta(t), \delta(t))$$=d_{\Vert 1}|\Vert((B^{-\frac{1}{2}}AB^{-z})^{1-t}, (B^{-\frac{1}{2}}CB^{-\frac{1}{2}})^{1-t})+d_{\Vert 1}1\Vert|((C^{-z}BC^{-f})^{t}, (C^{-q}DC^{-5})^{t})\iota 111$$\leqq(1-t)d_{\Uparrow 1}|\Downarrow(B^{-\frac{1}{2}}AB^{-\frac{1}{2}}, B^{-\frac{1}{2}}CB^{-1})+td_{1\Vert}|\Vert(C^{-l}fBC^{-\frac{1}{2}}, C^{-1}zDC^{-\frac{1}{2}})$.

$=(1-t)d_{\Vert 1}\Vert|(A, C)+td_{1\Vert}\Vert|(B, D)=(1-t)d_{\Vert 1}\Vert|(\gamma(0), \delta(0))+td_{|\Vert\Vert|}(\gamma(1), \delta(1))$

Let $\Gamma(t)=(Am_{p}B)m_{t}(Am_{q}B)$ and $\triangle(t)=(Cm_{p}D)m_{t}(Cm_{q}D)$ . Then, by theinterpolationality, we have$\Gamma(t)=Am(1-t)p+tqB$

and$\triangle(t)=Cm(1-t)p+tqD$ ,so that,$\Gamma(0)=Am_{p}B$ ,$\Gamma(1)=Am_{q}B$ ,Thus, we have (iii) for$\Delta(0)=Cm_{p}D$and$\Delta(1)=Cm_{q}D$.

and . Considering geodesics$\gamma(t)=A^{t}\in P(I, A)$ and $\delta(t)=B^{t}\in P(I, A)$ , we have (i) implies$\Gamma$and$\triangle$

implies (ii) for$\delta$$\gamma$$d_{\Vert|\Vert|}(A^{t}, B^{t})=F(t)=F((1-t)0+t)\leqq(1-t)F(O)+tF(1)$$=(1-t)d_{Y1}\Uparrow(I, I)+td_{\int 1}|\Vert(A, B)=td_{N1}M(A, B)$.

Thus all the relations hold and they are equivalent.

3

口Extreme Finsler geometry
Only two metric$d_{\pm 1}$

are derived from Finsler metrics since$A\nabla_{t}B=Am_{1,t}B=Am_{1_{2}t}B$and$A!_{t}B=Am_{-1,t}B=Am_{-1t\}}B$ .

Here we observe Finsler geometries and metrics for $r=\pm 1$ . First we see the trivial
fiber bundle
where is the unitary group of a unital -algebra$(\mathcal{A}^{+}\cross \mathcal{U}, \mathcal{A}^{+},\mathcal{U}, \pi_{1})$$\mathcal{U}$$C^{*}$

35$\mathcal{A}$where $\pi_{1}((A, U))=-4$ and the action is $L_{(A,U)}B=UBU$ “. Then the fiber isand the horizontal lift of a curve is $\Gamma(t)=(\gamma/(t). U)$ for a fixed. So the parallel displacement
and the covariant derivative are trivial:$\pi^{-1}(A)=A\cross \mathcal{U}$$\Gamma$$\gamma$$U\in \mathcal{U}$$P_{t}$

and$P_{t}X=X$$D_{t}X(t)= \frac{dX}{dt}(t)$.

Thereby the geodesic equation is. Thus we have the geodesic from$A$
to $B$ is the arithmetic mean $A\nabla_{t}B=(1-t)A+tB$ . Here we define a trivial
Finsler metric $L_{1}(X;A)=\Vert|X\Vert|$ and then$0=D_{t}\dot{\gamma}=\ddot{\gamma}(t)$$\int_{0^{L_{1}(\wedge}}^{1}/(t);\gamma(t)dt=\int_{0}^{1}\Vert|\dot{\gamma}(t)\Vert|dt=\int_{0}^{1}\Vert|B-A\Vert|dt=\Vert|B-A\Vert|$

We can verify$A\nabla_{t}B$.

attains the shortest: For any $\gamma\in P(A, B)$ , we have.

$\int_{0}^{1}L_{1}(\dot{\gamma}(t);\gamma(t))dt\geqq\Vert|\int_{0}^{1}\dot{\gamma}(t)dt\Vert|=\Vert|[\gamma(t)]_{0}^{1}\Vert|=\Vert|B-A\Vert|=d_{1}(A, B)$

Remark 3. Like Theorem 5, if the norm is strictly convex, then the geodesic is aunique path attaining the shortest. Suppose the norm is the operator one or the
trace one, a path
also attains the shortest length for $B\geq I$ .
$B^{t}$

Next we consider the trivial bundle
which is the same as thepreceding case except the action $L_{(A.U)}B=AUBU^{*}A$ . Then the parallel displacement is$(\mathcal{A}^{+}\cross \mathcal{U}, \mathcal{A}^{+},\mathcal{U}, \pi_{1})$$P_{t}X=\gamma(t)U((\gamma(0)U)^{-1}X(U^{*}\gamma(0))^{-1})U^{*}\gamma(t)=\gamma(t)\gamma(0)^{-1}X\gamma(0)^{-1}\gamma(t)$and hence the covariant derivative is obtained by$D_{t}(X(t))= \lim_{\epsilonarrow 0}\frac{1}{\epsilon}(\gamma(t)\gamma(t+\epsilon)^{-1}X(t+\epsilon)\gamma(t+\epsilon)^{-1}\gamma(t)-X(t))$$=\gamma(\gamma^{-1}X\gamma^{-1})’\gamma(t)$$=\gamma(-\gamma^{-1}\dot{\gamma}\gamma^{-1}X\gamma(t)^{-1}+\gamma^{-1}\dot{X}\gamma^{-1}-\gamma^{-1}X\gamma^{-1}\dot{\gamma}\gamma^{-1})\gamma(t)$$=(\dot{X}-\dot{\gamma}\gamma^{-1}X-X\gamma^{-1}\dot{\gamma})(t)$.

The geodesic equation is$0=\ddot{\gamma}(t)-2\dot{\gamma}(t)\gamma^{-1}(t)\dot{\gamma}(t)$and thereby$(-\gamma^{-1}\dot{\gamma}\gamma^{-1})’=\gamma^{-1}\dot{\gamma}\gamma^{-1}\dot{\gamma}\gamma^{-1}-\gamma^{-1}\ddot{\gamma}\gamma^{-1}+\gamma^{-1}\dot{\gamma}\gamma^{-1}\dot{\gamma}\gamma^{-\cdot 1}$.

$=2\gamma^{-1}\dot{\gamma}\gamma^{-1}\dot{\gamma}\gamma^{-1}-\gamma^{-1}\ddot{\gamma}\gamma^{-1}=\gamma^{-1}(2\dot{\gamma}\gamma^{-1}\dot{\gamma}-\ddot{\gamma})\gamma^{-1}=0$

36
So there exists a seif-adjoinr operator

$C$

with

$C=-\gamma\prime^{-1}(t),\wedge(t)\gamma^{-1}(t)=(\gamma^{-1}(t))’$and consequently. $D+tC=\gamma^{-1}(t)$ for some positive operator $D$ . Since $A=\gamma(O)=$. $B=\gamma(1)=(D+C)^{-1}$ , we have $D=A^{-1},$ $C=B^{-1}-D=B^{-1}-A^{-1,}$. that
$D^{-1}$is,$\gamma(t)=(A^{-1}+t(B^{-1}-A^{-1}))^{-1}=((1-t)A^{-1}+tB^{-1})^{-1}=A!_{t}B$ .

Define $L_{-1}(X;A)=\Vert|A^{-1}XA^{-1}\Vert|$ . Then it is a Finsler metric. In fact,$L_{-1}(P_{t}X|\gamma(t))=\Vert|I\gamma(0)^{-1}X\gamma(0)^{-1}I\Vert|=\Vert|\gamma(0)^{-1}X\gamma(0)^{-1}\Vert|=L_{-1}(X;\gamma(0))$

Now, for$f(t)=\gamma^{\prime^{-)}}$.

, we have

タ $=(f^{-1})’=-f^{-1}ff^{-1}=-\gamma\cdot(B^{-1}-A^{-1})\cdot\gamma$ .

Therefore$\int_{0}^{1}L_{-1}(\dot{\gamma}(t);\gamma(t)dt=\int_{0}^{1}\Vert|-\gamma^{-1}\gamma\cdot(B^{-1}-A^{-1})\cdot\gamma\gamma^{-1}\Vert|dt$$= \int_{0}^{1}\Vert|B^{-1}-A^{-1}\Vert|dt=\Vert|B^{-1}-A^{-1}\Vert|$,

and moreover it attains the shortest:$\int_{0}^{1}L_{-1}(\dot{\gamma};\gamma)dt\geqq\Vert|\int_{0}^{1}\gamma^{-1}\dot{\gamma}\gamma^{-1}dt\Vert|=\Vert|-[\gamma^{-1}]_{0}^{1}\Vert|=\Vert|B^{-1}-A^{-1}\Vert|$.

Remark 4. Like Theorem 5, if the norm is strictly convex, then the geodesic is aunique path attaining the shortest. Suppose the norm is the operator one or the
trace one, a path
also attains the shortest length for $B\leq I$ .
$B^{t}$

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