Let $f: G= \mbox{GL}(n,\mathbb{R}) \to \mathbb{R}$ be the determinant function. Then $\mbox{Hess} (f)$ is a two linear map on $M_{n}(\mathbb{R})\simeq T_{e}(G)$ where $e$ is the neutral element of $G$, the identity matrix. What is an explicit formula for this Hessian? (In terms of matrix terminologies)

3$\begingroup$ I think you can find what you're looking for by differentiating Jacobi's formula. Not sure if the answer will be nicely expressible in terms of standard matrix operations, though. $\endgroup$– David ZhangSep 6, 2016 at 22:59

$\begingroup$ @DavidZhang Thanks for your comment. $\endgroup$– Ali TaghaviSep 6, 2016 at 23:15
1 Answer
The formula you're looking for can be obtained by differentiating Jacobi's formula $$ \frac{\mathrm{d}}{\mathrm{d}t} \det A(t) = \det A(t) \cdot \operatorname{tr}\left( A^{1} \frac{\mathrm{d}A}{\mathrm{d}t} \right) $$ with respect to a second parameter, say $s$: \begin{multline} \frac{\partial^2}{\partial s \partial t} \det A(s,t) = \det A(s,t) \cdot \bigg[ \operatorname{tr}\left( A^{1} \frac{\partial A}{\partial s} \right) \operatorname{tr}\left( A^{1} \frac{\partial A}{\partial t} \right) \\ + \operatorname{tr}\left( A^{1} \frac{\partial^2 A}{\partial s \partial t} \right)  \operatorname{tr}\left( A^{1} \frac{\partial A}{\partial s} A^{1} \frac{\partial A}{\partial t} \right) \bigg] \end{multline} Now take $s = A_{ij}$ and $t = A_{kl}$, so $\frac{\partial A}{\partial s} = E_{ij}$ is the matrix with a one in its $(i,j)$entry, and zeros elsewhere. Similarly $\frac{\partial A}{\partial t} = E_{kl}$, and $\frac{\partial^2 A}{\partial s \partial t} = 0$. The desired Hessian is then \begin{align*} \operatorname{Hess}(\det)_A(U,V) &= U_{ij} V_{kl} (\det A)\bigg[ \operatorname{tr}\left( A^{1} E_{ij} \right) \operatorname{tr}\left( A^{1} E_{kl} \right)  \operatorname{tr}\left( A^{1} E_{ij} A^{1} E_{kl} \right) \bigg] \\ &= U_{ij} V_{kl} (\det A)\bigg[ (A^{1})_{mn} (E_{ij})_{nm} (A^{1})_{pq} (E_{kl})_{qp} \\ &\hspace{4cm}  (A^{1})_{mn} (E_{ij})_{np} (A^{1})_{pq} (E_{kl})_{qm} \bigg] \\ &= U_{ij} V_{kl} (\det A)\bigg[ (A^{1})_{mn} \delta_{in} \delta_{jm} (A^{1})_{pq} \delta_{kq} \delta_{lp} \\ &\hspace{4cm}  (A^{1})_{mn} \delta_{in} \delta_{jp} (A^{1})_{pq} \delta_{kq} \delta_{lm} \bigg] \\ &= U_{ij} V_{kl} (\det A)\bigg[ (A^{1})_{ji} (A^{1})_{lk}  (A^{1})_{li} (A^{1})_{jk} \bigg] \\ &= \det A \bigg[ U_{ij} (A^{1})_{ji} V_{kl} (A^{1})_{lk}  U_{ij} (A^{1})_{jk} V_{kl} (A^{1})_{li} \bigg] \\ &= \det A \bigg[ \operatorname{tr}(U A^{1}) \operatorname{tr}(V A^{1})  \operatorname{tr}(U A^{1} V A^{1}) \bigg] \\ \end{align*} with the Einstein summation convention in full force throughout. By evaluating this formula at $A = e,$ the $n \times n $ identity matrix, we obtain the desired special case $$ \operatorname{Hess}(\det)_e(U,V) = \operatorname{tr}(U) \operatorname{tr}(V)  \operatorname{tr}(U V). $$


$\begingroup$ Thank you again for your computations and very interesting formula. The formula is invariant under the natural action of $G$ on $\mathfrak{g}\times \mathfrak{g}$.Do you think that $\endgroup$ Sep 12, 2016 at 8:00

$\begingroup$ ....it is natural in the following sense: If a function is constant on every conjugacy class of a Lie group, then its Hessian at e is invariant under the natural action of $G$ on $\mathfrak{g} \times \mathfrak {g}$? $\endgroup$ Sep 12, 2016 at 8:04

$\begingroup$ @AliTaghavi I am not sure. Perhaps you can ask this as a separate question. $\endgroup$ Sep 12, 2016 at 16:19

$\begingroup$ Yes I try to ask it. BTW, your formula define a pseudo Riemannian metric on $Gl(n, \mathbb{R}$, since the two form is non degenerate.Even more: $SO(n)$ is seudo Riemannian submanifold. Is this special structure studied already?What can be said about the signature of this metric?Is $SO(n)$ a complet submanifold?Are you interested in such type of questions? $\endgroup$ Sep 12, 2016 at 18:48