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If we consider $G$ a compact Lie group, there is a left invariant Riemannian metric whose the sectional curvature is nonnegative (see Milnors' paper). When can we find a left invariant metric that has positive sectional curvature?

The unique simply connected Lie group with a left-invariant metric that is positively curved is $SU(2)$. This is in Milnor's paper cited by @Igor Rivin.

Is $\mathrm{SU}(2)$ the only positively curved simply connected Lie group (not necessarily with an invariant metric)?

How could we distinguish the positively curved Lie groups?

Are there obstructions in the case of Lie groups? The fact that $G$ is a Lie group could imply the existence of some good invariant of positively curved manifolds?

Remark: I made many edits on the question because some ideas are clearer now with help of the time and the answers and comments posted here. I would like to observe that I started with a question with a pseudo-conjecture that the only positively curved Lie groups are $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$ and asked for references.

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  • $\begingroup$ A mathscinet search on "left-invariant" and "nonnegative curvature" yields substantial info on the classification and properties of such metrics. $\endgroup$
    – Igor Belegradek
    Commented Nov 28, 2015 at 23:15
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    $\begingroup$ I find the question quite vague, for instance I don't see what "(non necessarily)" refers to. Also the last sentence is not very understandable, the information should be "It follows from the Bonnet-Myers theorem that any Lie group of dimension $\ge 2$ with a left-invariant Riemannian metric of positive sectional curvature is compact and has a finite fundamental group". $\endgroup$
    – YCor
    Commented Nov 29, 2015 at 0:11
  • $\begingroup$ There are currently 8 question marks here, which is too many, making it hard to see which question is being answered by which response. $\endgroup$
    – user44143
    Commented Nov 29, 2018 at 13:08
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    $\begingroup$ I don't think your first sentence is true. That is, not all left-invariant metrics on Lie groups are non-negatively curved, even in the compact simply connected case. Namely, the Berger metrics on $S^3$ (obtained by scaling the usual round metric in the direction of the Hopf fibers) has $2$-planes of negative curvature if you make the Hopf circles too large. $\endgroup$
    – Jason DeVito - on hiatus
    Commented Dec 4, 2018 at 14:28

2 Answers 2

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The following result is, for example, exercise 3 on pg. 104 of Do Carmo's Riemannian Geometry book.

Suppose $X$ is a Killing field on a compact even dimensional Riemannian manifold of positive curvature. Then $X$ has a zero.

Using this result, it's very easy to prove the following generalization of Wallach's theorem.

Theorem: No compact Lie group $G$ of rank $2$ or higher admits a positively curved Riemannian metric which is invariant under left multiplication by any $T^2\subseteq G$.

Proof: Suppose there is such a group $G$. Because the $T^2$ action on $G$ by left multiplication is free, the action fields associated to it have no zeros. On the other hand, because the $T^2$ is isometric, the action fields are are Killing fields. By the above result, this implies the dimension of $G$ is odd.

Because the action of $S^1\times \{1\}\subseteq T^2$ on $G$ is isometric, the even dimensional manifold $G/(S^1\times \{1\})$ inherits a Riemannian metric for which the action by $\{1\}\times S^1$ is free and isometric. Further, by O'Neill's formulas for a Riemannian submersion, the induced metric on $G/(S^1\times \{1\})$ is positively curved. But then, just as above, the action field is non-zero and Killing, contradicting the quoted result above. $\square$

Incidentally, without some kind of symmetry assumption, we can say basically nothing about the existence of positively curved metrics on compact Lie groups with finite $\pi_1$. The issue that among closed simply connected manifolds, there is no known obstruction which distinguishes those that admits metrics of non-negative curvature from those which admit metrics of positive curvature.

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    $\begingroup$ Do you mean for instance that it's unknown whether $S^3\times S^3$ admits a positively curved metric? $\endgroup$
    – YCor
    Commented Nov 30, 2016 at 5:24
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    $\begingroup$ @YCor: Yep, completely unknown. That said, there are some conjectures which imply the answer is no. (For example, a version of a conjecture of Bott asserts that an even dimensional positively curved manifold should have positive Euler characteristic.) $\endgroup$
    – Jason DeVito - on hiatus
    Commented Nov 30, 2016 at 13:17
  • $\begingroup$ @JasonDeVito: Isn't the above conjecture due to Hopf about Euler characteristic? $\endgroup$
    – C.F.G
    Commented Dec 4, 2018 at 4:35
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    $\begingroup$ @C.F.G. I agree it's a conjecture of Hopf. Not sure why I wrote Bott there (though there is an important conjecture due to Bott regarding non-negatively curved simply connected compact manifolds) $\endgroup$
    – Jason DeVito - on hiatus
    Commented Dec 4, 2018 at 14:25
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The standard reference is:

Curvatures of left invariant metrics on lie groups by J Milnor - ‎1976

In particular, a theorem of Wallace (mentioned in Milnor's paper) confirms your conjecture.

(PS: The paper seems to be available to all).

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  • $\begingroup$ Do you know if it is true for general metrics or only for left-invariants? Or this question is so hard like Hopf's conjecture? $\endgroup$
    – melomm
    Commented Nov 28, 2015 at 19:23
  • $\begingroup$ @user2304 The theorem is for left-invariant metrics - I expect that it might be hard in general (in particular, $SU(2)\times SU(2)$ is pretty close to the Hopf conjecture...) $\endgroup$
    – Igor Rivin
    Commented Nov 28, 2015 at 19:26
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    $\begingroup$ Since there's no conjecture in the question, I'm wondering what you call "your conjecture"... after several minutes I finally guessed that you mean that the only positively curved Lie groups (of dimension $\ge 2$ and with left-invariant Riemannian metric) are $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$? $\endgroup$
    – YCor
    Commented Nov 29, 2015 at 0:13
  • $\begingroup$ @YCor thanks you formulated better my question. I'll make a edition in my question and put your version of the question. $\endgroup$
    – melomm
    Commented Nov 29, 2015 at 16:27
  • $\begingroup$ @YCor Yes, that is correct :) $\endgroup$
    – Igor Rivin
    Commented Nov 29, 2015 at 16:47

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