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Does there exist a bi-Invariant metric on $SL_n(\mathbb{R})$. I tried to google a bit but I didn't find anything helpful.

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  • $\begingroup$ Doesn' t the negative of the Killing form do? $\endgroup$ – Mariano Suárez-Álvarez Dec 20 '10 at 16:04
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    $\begingroup$ Ah: That gives a biinvariant semiRiemannian metric, rather. $\endgroup$ – Mariano Suárez-Álvarez Dec 20 '10 at 16:06
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A connected Lie group has a biinvariant metric iff it is isomorphic to the dirtect product of a compact one and a vector space---see Milnor's Curvatures of left invariant metrics. Such a factorization would carry over to the Lie algebras, etc.

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It's enough to show the answer is "no" for $SL_2(\mathbb{R})$, because $SL_2(\mathbb{R})$ is naturally embedded in $SL_n(\mathbb{R})$, and if $SL_n(\mathbb{R})$ has a biinvariant metric, then the induced metric on $SL_2(\mathbb{R})$ would be biinvariant.

So let's focus on $SL_2(\mathbb{R})$. Finding a biinvariant metric on $SL_2(\mathbb{R})$ is equivalent to finding an $Ad(SL_2(\mathbb{R})$ invariant inner product on $\mathfrak{sl_2(\mathbb{R})}$, so I'll show there is no such $Ad$ invariant metric.

Now, $\mathfrak{sl_2(\mathbb{R})}$ is 3 dimensional with basis $E_{12}, E_{21}, E_{11} - E_{22}$ where $E_{ij}$ denotes the matrix with a 1 in the $ij$ slot and a 0 elsewhere.

Consider $A = diag(2,1/2)$ ,the diagonal matrix with diagonal entries 2 and 1/2. This is clearly in $SL_2(\mathbb{R})$. Note then that $AE_{12}A^{-1} = 4 E_{12}$.

But if an inner product were $Ad$ invariant, it would have to preserve the length of $E_{12}$, giving a contradiction.

Thus, there is no $Ad$ invariant inner product on $\mathfrak{sl_2(\mathbb{R})}$, and hence there is no biinvariant metric on $SL_n(\mathbb{R})$.

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  • $\begingroup$ Can you please explain why a left (resp. right) invariant Riemannian distance function must corresponded to a left (resp. right) invariant Riemannian metric (inner product)? $\endgroup$ – Insubordinate Jul 23 '19 at 7:13
  • $\begingroup$ Within Lie theory, "bi-invariant metric" typically means "bi-invariant Riemannian metric"; I wrote the above answer under that assumption. It is not clear to me one way or the other whether every invariant distance function comes from an invariant Riemannian metric. On $\mathbb{R}^n$, it's true, but I don't know about a general Lie group. $\endgroup$ – Jason DeVito Jul 23 '19 at 14:28
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If $d$ were a bi-invariant metric on $\operatorname{SL}_n(\mathbb{R})$, we would be able to restrict it to a bi-invariant metric on $\operatorname{SL}_2(\mathbb{R})$ as in Jason's answer.

And then we would have $$ d\left(\left[{\begin{array}{cc} t & 0 \\ 0 & 1/t \\ \end{array}}\right]\left[{\begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array}}\right]\left[{\begin{array}{cc} 1/t & 0 \\ 0 & t \\ \end{array}}\right], \left[{\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array}}\right]\right)= d\left(\left[{\begin{array}{cc} 1 & t^2 \\ 0 & 1 \\ \end{array}}\right], \left[{\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array}}\right]\right)\to 0$$

as $t\to 0$. On the other hand, we also have

$$ d\left(\left[{\begin{array}{cc} t & 0 \\ 0 & 1/t \\ \end{array}}\right]\left[{\begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array}}\right]\left[{\begin{array}{cc} 1/t & 0 \\ 0 & t \\ \end{array}}\right], \left[{\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array}}\right]\right)= d\left( \left[{\begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array}}\right], \left[{\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array}}\right]\right) \neq 0. $$

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