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Cross post from MSE. and sorry if this is an obvious question.

Here is a line of proof of Theorem 1.15 from

Brendle, Simon, Ricci Flowflow and the Sphere Theorem by Simon Brendlesphere theorem, Graduate Studies in Mathematics 111. Providence, RI: American Mathematical Society (AMS) (ISBN 978-0-8218-4938-5/hbk). vii, 176 p. (2010). ZBL1196.53001.

Let us fix two points $p, q \in M$ such that $d(p, q) = \operatorname{diam}(M, g) > 2$$d(p, q) = \operatorname{diam}(M, g) > \frac{\pi}{2}$. Since $\pi_k(M)\neq 0$, there exists a geodesic $\gamma : [0,1] \to M$ such that $\gamma(0) = \gamma(1) = p$ and $\operatorname{ind}(\gamma) < k$.

Q: Why $\pi_k(M)\neq 0 \implies \operatorname{ind}(\gamma) < k$? Is this a general fact?

Note: $\operatorname{ind}(\gamma):=$ Morse index of $\gamma$.

Cross post from MSE. and sorry if this is an obvious question.

Here is a line of proof of Theorem 1.15 from Ricci Flow and the Sphere Theorem by Simon Brendle

Let us fix two points $p, q \in M$ such that $d(p, q) = \operatorname{diam}(M, g) > 2$. Since $\pi_k(M)\neq 0$, there exists a geodesic $\gamma : [0,1] \to M$ such that $\gamma(0) = \gamma(1) = p$ and $\operatorname{ind}(\gamma) < k$.

Q: Why $\pi_k(M)\neq 0 \implies \operatorname{ind}(\gamma) < k$? Is this a general fact?

Note: $\operatorname{ind}(\gamma):=$ Morse index of $\gamma$.

Cross post from MSE. and sorry if this is an obvious question.

Here is a line of proof of Theorem 1.15 from

Brendle, Simon, Ricci flow and the sphere theorem, Graduate Studies in Mathematics 111. Providence, RI: American Mathematical Society (AMS) (ISBN 978-0-8218-4938-5/hbk). vii, 176 p. (2010). ZBL1196.53001.

Let us fix two points $p, q \in M$ such that $d(p, q) = \operatorname{diam}(M, g) > \frac{\pi}{2}$. Since $\pi_k(M)\neq 0$, there exists a geodesic $\gamma : [0,1] \to M$ such that $\gamma(0) = \gamma(1) = p$ and $\operatorname{ind}(\gamma) < k$.

Q: Why $\pi_k(M)\neq 0 \implies \operatorname{ind}(\gamma) < k$? Is this a general fact?

Note: $\operatorname{ind}(\gamma):=$ Morse index of $\gamma$.

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C.F.G
  • 4.2k
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  • 65

Cross post from MSE. and sorry if this is aan obvious question.

Here is a line of proof of Theorem 1.15 from Ricci Flow and the Sphere Theorem by Simon Brendle

Let us fix two points $p, q \in M$ such that $d(p, q) = \operatorname{diam}(M, g) > 2$. Since $\pi_k(M)\neq 0$, there exists a geodesic $\gamma : [0,1] \to M$ such that $\gamma(0) = \gamma(1) = p$ and $\operatorname{ind}(\gamma) < k$.

Q: Why $\pi_k(M)\neq 0 \implies \operatorname{ind}(\gamma) < k$? Is this a general fact?

Note: $\operatorname{ind}(\gamma):=$ Morse index of $\gamma$.

Cross post from MSE. and sorry if this is a obvious question.

Here is a line of proof of Theorem 1.15 from Ricci Flow and the Sphere Theorem by Simon Brendle

Let us fix two points $p, q \in M$ such that $d(p, q) = \operatorname{diam}(M, g) > 2$. Since $\pi_k(M)\neq 0$, there exists a geodesic $\gamma : [0,1] \to M$ such that $\gamma(0) = \gamma(1) = p$ and $\operatorname{ind}(\gamma) < k$.

Q: Why $\pi_k(M)\neq 0 \implies \operatorname{ind}(\gamma) < k$? Is this a general fact?

Note: $\operatorname{ind}(\gamma):=$ Morse index of $\gamma$.

Cross post from MSE. and sorry if this is an obvious question.

Here is a line of proof of Theorem 1.15 from Ricci Flow and the Sphere Theorem by Simon Brendle

Let us fix two points $p, q \in M$ such that $d(p, q) = \operatorname{diam}(M, g) > 2$. Since $\pi_k(M)\neq 0$, there exists a geodesic $\gamma : [0,1] \to M$ such that $\gamma(0) = \gamma(1) = p$ and $\operatorname{ind}(\gamma) < k$.

Q: Why $\pi_k(M)\neq 0 \implies \operatorname{ind}(\gamma) < k$? Is this a general fact?

Note: $\operatorname{ind}(\gamma):=$ Morse index of $\gamma$.

Source Link
C.F.G
  • 4.2k
  • 6
  • 31
  • 65

Does $\pi_k(M)\neq 0$ implies $\operatorname{ind}(\gamma) < k$?

Cross post from MSE. and sorry if this is a obvious question.

Here is a line of proof of Theorem 1.15 from Ricci Flow and the Sphere Theorem by Simon Brendle

Let us fix two points $p, q \in M$ such that $d(p, q) = \operatorname{diam}(M, g) > 2$. Since $\pi_k(M)\neq 0$, there exists a geodesic $\gamma : [0,1] \to M$ such that $\gamma(0) = \gamma(1) = p$ and $\operatorname{ind}(\gamma) < k$.

Q: Why $\pi_k(M)\neq 0 \implies \operatorname{ind}(\gamma) < k$? Is this a general fact?

Note: $\operatorname{ind}(\gamma):=$ Morse index of $\gamma$.