Defining the inner product - 1/2 tr(XY) on the lie algebra so(n)(skew symmetric matrices) is one way to introduce a bi-invariant metric on So(n) since the inner product is ad-invariant. Are there any other non-trivial differentiable bi-invariant metrics on So(n)? Also, is the above defined metric equivalent to the metric induced on So(n) when it is seen as a fiber bundle over So(n-1)? Any inputs will be helpful.
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$\begingroup$ What map $So(n) \to So(n-1)$ are you talking about here? $\endgroup$– Will SawinCommented Apr 3, 2013 at 18:30
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$\begingroup$ Exact map is slightly tedious to write but the quotient So(n)/So(n-1) is (n-1) sphere and we can use it to write So(n) as a fiber bundle over So(n-1) where the fiber is the (n-1) sphere. Using this, we can inductively define a metric on So(n). $\endgroup$– sreedharCommented Apr 3, 2013 at 18:39
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$\begingroup$ sreedhar: $SO(n)$ fibers over $S^n$ with fiber $SO(n-1)$, not the other way around. Just think about trying to fiber $SO(3)$ over $SO(2)=S^1$. $\endgroup$– MishaCommented Apr 3, 2013 at 18:45
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3$\begingroup$ When $n=4$, there is a $2$-parameter family of bi-invariant metrics on $\mathrm{SO}(4)$, but for $n\not=4$, the bi-invariant Riemannian metric on $\mathrm{SO}(n)$ is unique up to a constant multiple. $\endgroup$– Robert BryantCommented Apr 3, 2013 at 20:11
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2$\begingroup$ @GB: Yes. The point is that, for $n\not=2,4$, the Lie algebra ${\frak{so}}(n)$ is simple. In particular, the representation $\mathrm{Ad}:\mathrm{SO}(n)\to \mathrm{Aut}\bigl({\frak{so}}(n)\bigr)$ is irreducible, so there is, up to multiples, only one $\mathrm{Ad}$-invariant quadratic form on ${\frak{so}}(n)$, and hence, up to multiples, only one bi-invariant Riemannian metric on $\mathrm{SO}(n)$ when $n\not=4$. However, because ${\frak{so}}(4)\simeq {\frak{so}}(3)\oplus{\frak{so}}(3)$, there is a $2$-parameter family of $\mathrm{Ad}$-invariant quadratic forms on ${\frak{so}}(4)$. $\endgroup$– Robert BryantCommented Apr 3, 2013 at 21:59
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