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Can there exist a right invariant killing field of a right invariant (but not bi-invariant) Riemannian metric on a Lie group?

I am especially interested in the case of $SU(N)$ with a metric of the form (at the identity): $g(x,y) = \frac{1}{\lambda} B(x,y) + \frac{1}{\lambda^2} B(x,w)B(y,w)$ where $w$ is an arbitrary given vector in $\mathfrak{su}(n)$ s.t. B(w,w) < 1 and B is the Killing form (taken to be positive definite).

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  • $\begingroup$ The answer to your question is trivially yes, because a bi-invariant metric is right-invariant. $\endgroup$
    – José Figueroa-O'Farrill
    Commented May 14, 2014 at 18:51
  • $\begingroup$ I should have clarified, right invariant but not bi-invariant. Sorry for the confusion. $B$ is bi-invariant but $w$ is only right invariant. As far as I know such a metric cannot be bi-invariant as $w$ would need to be bi-invariant and there are no such vector fields on $SU(n)$. $\endgroup$
    – Benjamin
    Commented May 14, 2014 at 18:52
  • $\begingroup$ Indeed. But I’m just pointing out, perhaps unhelpfully, that your first question as written has a trivial answer. Hence, perhaps you’d like to ask a more precise question. $\endgroup$
    – José Figueroa-O'Farrill
    Commented May 14, 2014 at 18:56

2 Answers 2

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A right invariant Riemannian metric is invariant und all right translations. A left invariant vector field has a flow consisting of right translations (by $\exp(tX)$). Thus each left invariant field is a Killing field. A right invariant field $R_X$ is Killing if and only if $S^2(\text{ad}_X)^*g_{e}=0$.

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  • $\begingroup$ I am aware that all left invariant vector fields are killing. I have arrived at the equation $g([a,w], a+w) = 0$ $\forall a \in \mathfrak{su}(n)$ as a condition for the specific case I mentioned. Is this a special case of your equation above? $\endgroup$
    – Benjamin
    Commented May 14, 2014 at 18:25
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There is an obvious right-invariant vector field which leaves that metric invariant: the one which extends $w$. As Peter Michor mentions in his answer, the condition for a right-invariant vector field to be Killing is that it should preserve the inner product at the identity. In your example, and letting $b = \lambda^{-1} B$, and using the ad-invariance of the Killing form, $$ (\operatorname{ad}_z^* g)(x,y) = - b(x,[z,w])b(y,w) - b(x,w)b(y,[z,w]) $$ which vanishes for $z = w$.

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  • $\begingroup$ Would it be sufficient to solve $(ad^*_z g)(x,x) = 0$ rather than introducing y also? $\endgroup$
    – Benjamin
    Commented May 14, 2014 at 19:11
  • $\begingroup$ Sure, you can recover the general case by polarisation as usual. $\endgroup$
    – José Figueroa-O'Farrill
    Commented May 14, 2014 at 21:15
  • $\begingroup$ Ok, thanks. That is really helpful. Am I right in thinking that the proof in your answer would be correct for arbitrary right invariant $b$ rather than just the killing form? I mean for a metric of the form in the original question but where $B$ is an arbitrary right invariant metric? $\endgroup$
    – Benjamin
    Commented May 14, 2014 at 23:27
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    $\begingroup$ Yes, all I used was the ad-invariance of $b$. $\endgroup$
    – José Figueroa-O'Farrill
    Commented May 14, 2014 at 23:58

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