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Malkoun
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Question 3: is the image of $f$ diffeomorphic to $S^4$? If so, then it would provide yet another proof of the previous folklore result. A related question is whether or not $V$ is diffeomorphic to $S^4$. Edit: I think I can build a diffeomorphism from $S^4$ onto $V$. Think of $S^4$ as the set $$W = \{ B \,|\, \text{$B$ real symmetric $3$-by-$3$, } \operatorname{tr}(B) = 0 \text{ and } \operatorname{tr}(B^2) = 1 \}.$$ It is not too difficult to see that $W$ is diffeomorphic to $S^4$. Define a map from $W$ into $V$, by $$ B \mapsto \frac{B - \lambda_2(B) I}{\lVert B - \lambda_2(B)I \rVert},$$ where $\lambda_1(B) \leq \lambda_2(B) \leq \lambda_3(B)$ are the $3$ eigenvalues of $B$. I think that this map is perhaps a diffeomorphism from $W$ onto $V$. However, I am not sure about its smoothness when $2$ eigenvalues of $B$ collide. Can someone comment on that please?

Question 3: is the image of $f$ diffeomorphic to $S^4$? If so, then it would provide yet another proof of the previous folklore result. A related question is whether or not $V$ is diffeomorphic to $S^4$.

Question 3: is the image of $f$ diffeomorphic to $S^4$? If so, then it would provide yet another proof of the previous folklore result. A related question is whether or not $V$ is diffeomorphic to $S^4$. Edit: I think I can build a diffeomorphism from $S^4$ onto $V$. Think of $S^4$ as the set $$W = \{ B \,|\, \text{$B$ real symmetric $3$-by-$3$, } \operatorname{tr}(B) = 0 \text{ and } \operatorname{tr}(B^2) = 1 \}.$$ It is not too difficult to see that $W$ is diffeomorphic to $S^4$. Define a map from $W$ into $V$, by $$ B \mapsto \frac{B - \lambda_2(B) I}{\lVert B - \lambda_2(B)I \rVert},$$ where $\lambda_1(B) \leq \lambda_2(B) \leq \lambda_3(B)$ are the $3$ eigenvalues of $B$. I think that this map is perhaps a diffeomorphism from $W$ onto $V$. However, I am not sure about its smoothness when $2$ eigenvalues of $B$ collide. Can someone comment on that please?

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Gregory Arone
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I can answer question $2$. I edited accordingly.
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Malkoun
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$$V = \{ \det(A) = 0 \} \cap \{ \operatorname{tr}(A^2) = 1 \} \cap \{ \operatorname{tr}(A)^2 < 2 \}.$$$$V = \{ \det(A) = 0 \} \cap \{ \operatorname{tr}(A^2) = 1 \} \cap \{ \operatorname{tr}(A)^2 \leq 1 \}.$$

In other words, these conditions ensure that the eigenvalues of $A$, which must be real, are of the form: $0$, $\lambda$, $\mu$ with $\lambda \neq \mu$$\lambda \mu \leq 0$ and $\lambda^2 + \mu^2 = 1$ (note that $\lambda$, or $\mu$, may be $0$).

Question 2: is the image of $f$ equal to $V$? (IEdit: I think sothe image of $f$ is indeed $V$. Just note that it suffices to diagonalize $A$, and show that a diagonal matrix having $0$, $\lambda$ and $\mu$ as (real)? eigenvalues and satisfying the previous conditions is in the image of $f$. And this is straightforward.

$$V = \{ \det(A) = 0 \} \cap \{ \operatorname{tr}(A^2) = 1 \} \cap \{ \operatorname{tr}(A)^2 < 2 \}.$$

In other words, these conditions ensure that the eigenvalues of $A$, which must be real, are of the form: $0$, $\lambda$, $\mu$ with $\lambda \neq \mu$ and $\lambda^2 + \mu^2 = 1$ (note that $\lambda$, or $\mu$, may be $0$).

Question 2: is the image of $f$ equal to $V$ (I think so)?

$$V = \{ \det(A) = 0 \} \cap \{ \operatorname{tr}(A^2) = 1 \} \cap \{ \operatorname{tr}(A)^2 \leq 1 \}.$$

In other words, these conditions ensure that the eigenvalues of $A$, which must be real, are of the form: $0$, $\lambda$, $\mu$ with $\lambda \mu \leq 0$ and $\lambda^2 + \mu^2 = 1$ (note that $\lambda$, or $\mu$, may be $0$).

Question 2: is the image of $f$ equal to $V$? Edit: I think the image of $f$ is indeed $V$. Just note that it suffices to diagonalize $A$, and show that a diagonal matrix having $0$, $\lambda$ and $\mu$ as (real) eigenvalues and satisfying the previous conditions is in the image of $f$. And this is straightforward.

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Malkoun
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I corrected some statements, based on GregoryArone's comment.
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