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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 $G$ be a Lie group implies 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 of 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 Nov 28 '15 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 Nov 29 '15 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. $\endgroup$ – Matt F. Nov 29 '18 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 Dec 4 '18 at 14:28
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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 Nov 30 '16 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 Nov 30 '16 at 13:17
  • $\begingroup$ @JasonDeVito: Isn't the above conjecture due to Hopf about Euler characteristic? $\endgroup$ – C.F.G Dec 4 '18 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 Dec 4 '18 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 Nov 28 '15 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 Nov 28 '15 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 Nov 29 '15 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 Nov 29 '15 at 16:27
  • $\begingroup$ @YCor Yes, that is correct :) $\endgroup$ – Igor Rivin Nov 29 '15 at 16:47

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