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Just a suggestion on emphasis. I really like Mark's last comment on Sunday. So I suggest the following: Let $U \in \mathbb R^{(n^2 - n)/2}$ be the set of upper right triangles in any positive definite symmetric matrix $W$ with all 1's on the diagonal. I think $U$ is open and connected. Now, regard the matrix inverse of $W$ as mapping $U$ to $ \mathbb R^{(n^2 - n)/2},$ giving the upper triangle of some matrix, and of course ignoring the resulting diagonal.

My suggestion is that this may be a smooth bijection. Surjectivity is plausible as the determinant of $W$ may get arbitrarily close to 0, at the same time that several off-diagonal entries of $W$ are closer to $\pm 1$ than to 0. Useful ingredients could be the Cholesky decomposition. Meanwhile, as Brendan mentioned, if it is a bijection it at least must be a smooth diffeomorphism around every point.

Well, it works for $n=2,$ and looks very pretty for $n=3$ but i have not proved it yet.

Just a suggestion on emphasis. I really like Mark's last comment on Sunday. So I suggest the following: Let $U \subseteq \mathbb R^{(n^2 - n)/2}$ be the set of upper right triangles in any positive definite symmetric matrix $W$ with all 1's on the diagonal. I think $U$ is open and connected. Now, regard the matrix inverse of $W$ as mapping $U$ to $ \mathbb R^{(n^2 - n)/2},$ giving the upper triangle of some matrix, and of course ignoring the resulting diagonal.

My suggestion is that this may be a smooth bijection. Surjectivity is plausible as the determinant of $W$ may get arbitrarily close to 0, at the same time that several off-diagonal entries of $W$ are closer to $\pm 1$ than to 0. Useful ingredients could be the Cholesky decomposition. Meanwhile, as Brendan mentioned, if it is a bijection it at least must be a smooth diffeomorphism around every point.

Well, it works for $n=2,$ and looks very pretty for $n=3$ but i have not proved it yet.

Just a suggestion on emphasis. I really like Mark's last comment on Sunday. So I suggest the following: Let $U \in \mathbb R^{(n^2 - n)/2}$ be the set of upper right triangles in any positive definite symmetric matrix $W$ with all 1's on the diagonal. I think $U$ is open and connected. Now, regard the matrix inverse of $W$ as mapping $U$ to $ \mathbb R^{(n^2 - n)/2},$ giving the upper triangle of some matrix, and of course ignoring the resulting diagonal.

My suggestion is that this may be a smooth bijection. Surjectivity is plausible as the determinant of $W$ may get arbitrarily close to 0. Useful ingredients could be the Cholesky decomposition. Meanwhile, as Brendan mentioned, if it is a bijection it at least must be a smooth diffeomorphism around every point.

Well, it works for $n=2,$ and looks very pretty for $n=3$ but i have not proved it yet.

Just a suggestion on emphasis. I really like Mark's last comment on Sunday. So I suggest the following: Let $U \in \mathbb R^{(n^2 - n)/2}$ be the set of upper right triangles in any positive definite symmetric matrix $W$ with all 1's on the diagonal. I think $U$ is open and connected. Now, regard the matrix inverse of $W$ as mapping $U$ to $ \mathbb R^{(n^2 - n)/2},$ giving the upper triangle of some matrix, and of course ignoring the resulting diagonal.

My suggestion is that this may be a smooth bijection. Surjectivity is plausible as the determinant of $W$ may get arbitrarily close to 0, at the same time that several off-diagonal entries of $W$ are closer to $\pm 1$ than to 0. Useful ingredients could be the Cholesky decomposition. Meanwhile, as Brendan mentioned, if it is a bijection it at least must be a smooth diffeomorphism around every point.

Well, it works for $n=2,$ and looks very pretty for $n=3$ but i have not proved it yet.

Just a suggestion on emphasis. I really like Mark's last comment on Sunday. So I suggest the following: Let $U \in \mathbb R^{(n^2 - n)/2}$ be the set of upper right triangles in positive definite symmetric matrices with all 1's on the diagonal. I think $U$ is open and connected. Now, regard the matrix inverse as mapping $U$ to $ \mathbb R^{(n^2 - n)/2},$ giving the upper triangle of some matrix, and of course ignoring the resulting diagonal.

My suggestion is that this may be a smooth bijection. Useful ingredients could be the Cholesky decomposition. Meanwhile, as Brendan mentioned, if it is a bijection it at least must be a smooth diffeomorphism around every point.

Well, it works for $n=2,$ and looks very pretty for $n=3$ but i have not proved it yet.

Just a suggestion on emphasis. I really like Mark's last comment on Sunday. So I suggest the following: Let $U \in \mathbb R^{(n^2 - n)/2}$ be the set of upper right triangles in any positive definite symmetric matrix $W$ with all 1's on the diagonal. I think $U$ is open and connected. Now, regard the matrix inverse of $W$ as mapping $U$ to $ \mathbb R^{(n^2 - n)/2},$ giving the upper triangle of some matrix, and of course ignoring the resulting diagonal.

My suggestion is that this may be a smooth bijection. Surjectivity is plausible as the determinant of $W$ may get arbitrarily close to 0. Useful ingredients could be the Cholesky decomposition. Meanwhile, as Brendan mentioned, if it is a bijection it at least must be a smooth diffeomorphism around every point.

Well, it works for $n=2,$ and looks very pretty for $n=3$ but i have not proved it yet.