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Suppose $H$ is a complex Hilbert space and $\mathbb{P}H$ be its projective space. $d$ is the distance function on $\mathbb{P}H$ associted to the Fubini-Study metric on $\mathbb{P}H$.

In the proof of theorem 1 in Freed's paper , he wrote

Fix $L_1\neq L_2\in \mathbb{P}H$ and let $V$ be the $2$-dimensional space $L_1 + L_2 ⊂ H$. The unitary automorphism of $H = V ⊕V^⊥$ which is $+1$ on $V$ and $−1$ on $V^⊥$ induces an isometry of $\mathbb{P}H$ which has $\mathbb{P}V$ as a component of its fixed point set. It follows that $\mathbb{P}V$ is totally geodesic. Therefore, to compute $d(L_1, L_2)$ we are reduced to the case of the complex projective line with its Fubini-Study metric: the round $2$-sphere.

Here is my Question:

I think that what the author mean in the theorefore part is that $\mathbb{P}V$ is totally geodesic leads to that $d_{\mathbb{P}V}(L_1,L_2)=d(L_1,L_2)$, where $d_{\mathbb{P}V}$ is the distance function associated to the Fubini-Study metric on $\mathbb{P}V$, but I donnot know why, or I misunderstood the contexts? Really hope someone can help me with this.

  • Your interpretation the right one. The reason why is due to the following theorem : Let be a Riemannian manifold and any set of isometries of . Let be the set of points of which are left fixed by all elements of . Then each connected component of is a closed totally geodesic submanifold of . You might want to look at this : https://math.stackexchange.com/questions/55376/fixed-points-set-of-an-isometry and the reference they give. – Popyaitte Dec 09 '19 at 17:40
  • yeah, I know $\mathbb{P}V$ is totally geodesic, but I donnot understand why the "therefore " part is right. – Michael Jarrett Dec 10 '19 at 09:19

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