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We equip $\mathrm{GL}(n,\mathbb{R})$ and $\mathrm{O}(n)$ with their left-invariant metrics, whose restrictions to the corresponding neutral elements is the standard inner product $\mathrm{Trace}(AB^{\mathrm{tr}})$ of $M_n(\mathbb{R})$.

Let $f:\mathrm{GL}(n,\mathbb{R})\to \mathrm{O}(n)$ be the Gram-Schmidt map. This map is uniquely defined via Gram-schmidt process which is applied to the columns of a matrix.

Is $f$ a Riemannian submersion with respect to these Riemannian structures? If the answer is negative, can we change the Riemannian metrics of these spaces (not necessarily invariant under group operations) to have $f$ as a Riemannian submersion?

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    $\begingroup$ For example, with Gram-Schmidt one feature lacking in the undergraduate textbook versions so often seen is that they aren't equivariant with respect to the permutation action of column vectors. But it can be done equivariantly. $\endgroup$
    – Ryan Budney
    Commented Mar 25, 2019 at 15:26
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    $\begingroup$ I forget how people do the equivalent version in general, but this should work: take the simplest step of Gram-Schmidt where you slide the n-th vector into the orthogonal complement of the span of the first (n-1)-vectors. Think of that process as the solution to an ODE. Average the ODE over the $\Sigma_n$-action, and take the solution to that equivariant ODE. That should be the equivariant gram-schmidt. $\endgroup$
    – Ryan Budney
    Commented Mar 25, 2019 at 22:13
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    $\begingroup$ The ODE is zero on $O(n)$. If you think of your matrix $A \in GL(n, \mathbb R)$ as $n$ column vectors $\vec v_1, \cdots, \vec v_n$ then I suppose the ODE would be $\frac{\partial \vec v_i}{\partial t} = -proj_{\vec v_1, \cdots, \vec v_{i-1}, \vec v_{i+1}, \cdots, \vec v_n}(\vec v_i)$. This ODE has exponential decay, so you have to extend time to $+\infty$ to get Gram-Schmidt. I suspect there is a way to get a closed-form solution to this ODE, let me think about it for a second. $\endgroup$
    – Ryan Budney
    Commented Mar 26, 2019 at 19:05
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    $\begingroup$ Here is a slightly different ODE that behaves similarly -- in particular it's got a cute matrix-level description $A' = A(I-A^TA)$. Here $A \in GL(n,\mathbb R)$. A nice feature of this ODE is $O(n)$ is precisely the fixed-point set. $\endgroup$
    – Ryan Budney
    Commented Mar 26, 2019 at 19:49
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    $\begingroup$ That last ODE is perhaps be most pleasant. If you diagonalize $A^TA$ over the orthogonal group, it gives you a change of coordinates where you can write down a closed-form solution. It decays exponentially to an equivariant Gram-Schmidt. $\endgroup$
    – Ryan Budney
    Commented Mar 27, 2019 at 6:57

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