Mixing
matrixgroup is a differentiable submanifold of $M_4(mathbb{R})$
$begingroup$
My question is already existtent Prove that $mathcal{O}(3,1)$ is a submanifold of $mathcal{M}_4(mathbb{R})$ , but $I$ don't understand the hint and there is no answer (and the question there is not exactly the same question I have). Thus, I put my own question here :
Consider the group $G:={Min M_4(mathbb{R})mid M^{tr}PM=P}$, where $P=operatorname{diag}(1,1,1,-1)$.
The claim is that $G$ is a $6$-dim. (differentiable) submanifold of $M_4(mathbb{R})$.
What I have done so far: Let $fcolon M_4(mathbb{R})to M_4(mathbb{R})$, $$f(M)=M^{tr}PM-P.$$
Then $G=f^{-1}(0)$ and $f$ is continuously differentiable.
For each $Min G$, the map $Df(M)colon M_4(mathbb{R})to operatorname{Im }(Df(M))$ given by $Df(M)=M^{tr}P+PM$ has to be surjective onto a $10$-dimensional subspace of $M_4(mathbb{R})$.
However, how to check $dim operatorname{Im }(Df(M))=10$ (as elementary as possible)? After checking this I can conclude that $G$ is a $6$-dim. submanifold of $M_4(mathbb{R})$.
analysis multivariable-calculus differential-topology submanifold
asked Jan 11 at 9:50
TheAppliedTheoreticalOneTheAppliedTheoreticalOne
717
$endgroup$
add a comment |
$begingroup$
My question is already existtent Prove that $mathcal{O}(3,1)$ is a submanifold of $mathcal{M}_4(mathbb{R})$ , but $I$ don't understand the hint and there is no answer (and the question there is not exactly the same question I have). Thus, I put my own question here :
Consider the group $G:={Min M_4(mathbb{R})mid M^{tr}PM=P}$, where $P=operatorname{diag}(1,1,1,-1)$.
The claim is that $G$ is a $6$-dim. (differentiable) submanifold of $M_4(mathbb{R})$.
What I have done so far: Let $fcolon M_4(mathbb{R})to M_4(mathbb{R})$, $$f(M)=M^{tr}PM-P.$$
Then $G=f^{-1}(0)$ and $f$ is continuously differentiable.
For each $Min G$, the map $Df(M)colon M_4(mathbb{R})to operatorname{Im }(Df(M))$ given by $Df(M)=M^{tr}P+PM$ has to be surjective onto a $10$-dimensional subspace of $M_4(mathbb{R})$.
However, how to check $dim operatorname{Im }(Df(M))=10$ (as elementary as possible)? After checking this I can conclude that $G$ is a $6$-dim. submanifold of $M_4(mathbb{R})$.
analysis multivariable-calculus differential-topology submanifold
asked Jan 11 at 9:50
TheAppliedTheoreticalOneTheAppliedTheoreticalOne
717
$endgroup$
add a comment |
$begingroup$
My question is already existtent Prove that $mathcal{O}(3,1)$ is a submanifold of $mathcal{M}_4(mathbb{R})$ , but $I$ don't understand the hint and there is no answer (and the question there is not exactly the same question I have). Thus, I put my own question here :
Consider the group $G:={Min M_4(mathbb{R})mid M^{tr}PM=P}$, where $P=operatorname{diag}(1,1,1,-1)$.
The claim is that $G$ is a $6$-dim. (differentiable) submanifold of $M_4(mathbb{R})$.
What I have done so far: Let $fcolon M_4(mathbb{R})to M_4(mathbb{R})$, $$f(M)=M^{tr}PM-P.$$
Then $G=f^{-1}(0)$ and $f$ is continuously differentiable.
For each $Min G$, the map $Df(M)colon M_4(mathbb{R})to operatorname{Im }(Df(M))$ given by $Df(M)=M^{tr}P+PM$ has to be surjective onto a $10$-dimensional subspace of $M_4(mathbb{R})$.
However, how to check $dim operatorname{Im }(Df(M))=10$ (as elementary as possible)? After checking this I can conclude that $G$ is a $6$-dim. submanifold of $M_4(mathbb{R})$.
analysis multivariable-calculus differential-topology submanifold
asked Jan 11 at 9:50
TheAppliedTheoreticalOneTheAppliedTheoreticalOne
717
$endgroup$
My question is already existtent Prove that $mathcal{O}(3,1)$ is a submanifold of $mathcal{M}_4(mathbb{R})$ , but $I$ don't understand the hint and there is no answer (and the question there is not exactly the same question I have). Thus, I put my own question here :
Consider the group $G:={Min M_4(mathbb{R})mid M^{tr}PM=P}$, where $P=operatorname{diag}(1,1,1,-1)$.
The claim is that $G$ is a $6$-dim. (differentiable) submanifold of $M_4(mathbb{R})$.
What I have done so far: Let $fcolon M_4(mathbb{R})to M_4(mathbb{R})$, $$f(M)=M^{tr}PM-P.$$
Then $G=f^{-1}(0)$ and $f$ is continuously differentiable.
For each $Min G$, the map $Df(M)colon M_4(mathbb{R})to operatorname{Im }(Df(M))$ given by $Df(M)=M^{tr}P+PM$ has to be surjective onto a $10$-dimensional subspace of $M_4(mathbb{R})$.
However, how to check $dim operatorname{Im }(Df(M))=10$ (as elementary as possible)? After checking this I can conclude that $G$ is a $6$-dim. submanifold of $M_4(mathbb{R})$.
analysis multivariable-calculus differential-topology submanifold
analysis multivariable-calculus differential-topology submanifold
asked Jan 11 at 9:50
TheAppliedTheoreticalOneTheAppliedTheoreticalOne
717
asked Jan 11 at 9:50
TheAppliedTheoreticalOneTheAppliedTheoreticalOne
717
asked Jan 11 at 9:50
TheAppliedTheoreticalOneTheAppliedTheoreticalOne
717
asked Jan 11 at 9:50
TheAppliedTheoreticalOneTheAppliedTheoreticalOne
717
asked Jan 11 at 9:50
TheAppliedTheoreticalOneTheAppliedTheoreticalOne
717
717
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A more direct way to see this which has no need for any explicit calculations by hand: For every symmetric $A$ and $Y$, we can solve $MA+AM^T = Y$ for $M$ by finding any matrix $widetilde{Y}$, symmetric or otherwise, that satisfies $widetilde{Y}+widetilde{Y}^T = Y$ (for example $widetilde{Y}_{ij} := begin{cases} Y_{ij} & i<j \ frac{1}{2}Y_{ii} & i=j \ 0 & i>j end{cases}$ does it) and solving $MA = widetilde{Y}$ for $M$ instead. If $A$ is invertible, the latter is always solvable.
answered Jan 13 at 18:02
Johannes HahnJohannes Hahn
4,212634
$endgroup$
add a comment |
$begingroup$
The most elementary way consists in computing $Df_0(E_{ij})$ for each $i,jin{1,2,3,4}$, where $E_{ij}$ is the $4times4$ matrix such that the entry located at the column $i$ and the row $j$ is $1$ and all other entries are $0$. You shall comput $16$ matrices, each of which is symmetric. Then, check that the space that they span is the space of all $4times4$ symmetric matrices, whose dimension is $10$.
answered Jan 11 at 10:55
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
$begingroup$
I see. Thank you
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 11:00
$begingroup$
But you mean $Df(E_{ij})$, right?
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 15:47
$begingroup$
Actually, it's $Df_0$, that is, the derivative of $f$ at the null matrix. I've edited my answer.
$endgroup$
– José Carlos Santos
Jan 11 at 15:50
$begingroup$
ah ok, I just realized that $Df(E_{ij})$ is as well not correct, because the $E_{ij}$ are not in G
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 15:52
$begingroup$
Ok thanks, but why to consider the derivative of $f$ at the null matrix? I don't understand it
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 16:00
|
show 1 more comment
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A more direct way to see this which has no need for any explicit calculations by hand: For every symmetric $A$ and $Y$, we can solve $MA+AM^T = Y$ for $M$ by finding any matrix $widetilde{Y}$, symmetric or otherwise, that satisfies $widetilde{Y}+widetilde{Y}^T = Y$ (for example $widetilde{Y}_{ij} := begin{cases} Y_{ij} & i<j \ frac{1}{2}Y_{ii} & i=j \ 0 & i>j end{cases}$ does it) and solving $MA = widetilde{Y}$ for $M$ instead. If $A$ is invertible, the latter is always solvable.
answered Jan 13 at 18:02
Johannes HahnJohannes Hahn
4,212634
$endgroup$
add a comment |
$begingroup$
A more direct way to see this which has no need for any explicit calculations by hand: For every symmetric $A$ and $Y$, we can solve $MA+AM^T = Y$ for $M$ by finding any matrix $widetilde{Y}$, symmetric or otherwise, that satisfies $widetilde{Y}+widetilde{Y}^T = Y$ (for example $widetilde{Y}_{ij} := begin{cases} Y_{ij} & i<j \ frac{1}{2}Y_{ii} & i=j \ 0 & i>j end{cases}$ does it) and solving $MA = widetilde{Y}$ for $M$ instead. If $A$ is invertible, the latter is always solvable.
answered Jan 13 at 18:02
Johannes HahnJohannes Hahn
4,212634
$endgroup$
add a comment |
$begingroup$
A more direct way to see this which has no need for any explicit calculations by hand: For every symmetric $A$ and $Y$, we can solve $MA+AM^T = Y$ for $M$ by finding any matrix $widetilde{Y}$, symmetric or otherwise, that satisfies $widetilde{Y}+widetilde{Y}^T = Y$ (for example $widetilde{Y}_{ij} := begin{cases} Y_{ij} & i<j \ frac{1}{2}Y_{ii} & i=j \ 0 & i>j end{cases}$ does it) and solving $MA = widetilde{Y}$ for $M$ instead. If $A$ is invertible, the latter is always solvable.
answered Jan 13 at 18:02
Johannes HahnJohannes Hahn
4,212634
$endgroup$
A more direct way to see this which has no need for any explicit calculations by hand: For every symmetric $A$ and $Y$, we can solve $MA+AM^T = Y$ for $M$ by finding any matrix $widetilde{Y}$, symmetric or otherwise, that satisfies $widetilde{Y}+widetilde{Y}^T = Y$ (for example $widetilde{Y}_{ij} := begin{cases} Y_{ij} & i<j \ frac{1}{2}Y_{ii} & i=j \ 0 & i>j end{cases}$ does it) and solving $MA = widetilde{Y}$ for $M$ instead. If $A$ is invertible, the latter is always solvable.
answered Jan 13 at 18:02
Johannes HahnJohannes Hahn
4,212634
edited Jan 16 at 0:31
answered Jan 13 at 18:02
Johannes HahnJohannes Hahn
4,212634
answered Jan 13 at 18:02
Johannes HahnJohannes Hahn
4,212634
answered Jan 13 at 18:02
Johannes HahnJohannes Hahn
4,212634
4,212634
add a comment |
add a comment |
$begingroup$
The most elementary way consists in computing $Df_0(E_{ij})$ for each $i,jin{1,2,3,4}$, where $E_{ij}$ is the $4times4$ matrix such that the entry located at the column $i$ and the row $j$ is $1$ and all other entries are $0$. You shall comput $16$ matrices, each of which is symmetric. Then, check that the space that they span is the space of all $4times4$ symmetric matrices, whose dimension is $10$.
answered Jan 11 at 10:55
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
$begingroup$
I see. Thank you
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 11:00
$begingroup$
But you mean $Df(E_{ij})$, right?
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 15:47
$begingroup$
Actually, it's $Df_0$, that is, the derivative of $f$ at the null matrix. I've edited my answer.
$endgroup$
– José Carlos Santos
Jan 11 at 15:50
$begingroup$
ah ok, I just realized that $Df(E_{ij})$ is as well not correct, because the $E_{ij}$ are not in G
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 15:52
$begingroup$
Ok thanks, but why to consider the derivative of $f$ at the null matrix? I don't understand it
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 16:00
|
show 1 more comment
$begingroup$
The most elementary way consists in computing $Df_0(E_{ij})$ for each $i,jin{1,2,3,4}$, where $E_{ij}$ is the $4times4$ matrix such that the entry located at the column $i$ and the row $j$ is $1$ and all other entries are $0$. You shall comput $16$ matrices, each of which is symmetric. Then, check that the space that they span is the space of all $4times4$ symmetric matrices, whose dimension is $10$.
answered Jan 11 at 10:55
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
$begingroup$
I see. Thank you
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 11:00
$begingroup$
But you mean $Df(E_{ij})$, right?
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 15:47
$begingroup$
Actually, it's $Df_0$, that is, the derivative of $f$ at the null matrix. I've edited my answer.
$endgroup$
– José Carlos Santos
Jan 11 at 15:50
$begingroup$
ah ok, I just realized that $Df(E_{ij})$ is as well not correct, because the $E_{ij}$ are not in G
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 15:52
$begingroup$
Ok thanks, but why to consider the derivative of $f$ at the null matrix? I don't understand it
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 16:00
|
show 1 more comment
$begingroup$
The most elementary way consists in computing $Df_0(E_{ij})$ for each $i,jin{1,2,3,4}$, where $E_{ij}$ is the $4times4$ matrix such that the entry located at the column $i$ and the row $j$ is $1$ and all other entries are $0$. You shall comput $16$ matrices, each of which is symmetric. Then, check that the space that they span is the space of all $4times4$ symmetric matrices, whose dimension is $10$.
answered Jan 11 at 10:55
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
The most elementary way consists in computing $Df_0(E_{ij})$ for each $i,jin{1,2,3,4}$, where $E_{ij}$ is the $4times4$ matrix such that the entry located at the column $i$ and the row $j$ is $1$ and all other entries are $0$. You shall comput $16$ matrices, each of which is symmetric. Then, check that the space that they span is the space of all $4times4$ symmetric matrices, whose dimension is $10$.
answered Jan 11 at 10:55
José Carlos SantosJosé Carlos Santos
160k22126232
edited Jan 11 at 15:50
answered Jan 11 at 10:55
José Carlos SantosJosé Carlos Santos
160k22126232
answered Jan 11 at 10:55
José Carlos SantosJosé Carlos Santos
160k22126232
answered Jan 11 at 10:55
José Carlos SantosJosé Carlos Santos
160k22126232
160k22126232
$begingroup$
I see. Thank you
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 11:00
$begingroup$
But you mean $Df(E_{ij})$, right?
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 15:47
$begingroup$
Actually, it's $Df_0$, that is, the derivative of $f$ at the null matrix. I've edited my answer.
$endgroup$
– José Carlos Santos
Jan 11 at 15:50
$begingroup$
ah ok, I just realized that $Df(E_{ij})$ is as well not correct, because the $E_{ij}$ are not in G
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 15:52
$begingroup$
Ok thanks, but why to consider the derivative of $f$ at the null matrix? I don't understand it
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 16:00
|
show 1 more comment
$begingroup$
I see. Thank you
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 11:00
$begingroup$
But you mean $Df(E_{ij})$, right?
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 15:47
$begingroup$
Actually, it's $Df_0$, that is, the derivative of $f$ at the null matrix. I've edited my answer.
$endgroup$
– José Carlos Santos
Jan 11 at 15:50
$begingroup$
ah ok, I just realized that $Df(E_{ij})$ is as well not correct, because the $E_{ij}$ are not in G
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 15:52
$begingroup$
Ok thanks, but why to consider the derivative of $f$ at the null matrix? I don't understand it
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 16:00
$begingroup$
I see. Thank you
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 11:00
$begingroup$
I see. Thank you
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 11:00
$begingroup$
But you mean $Df(E_{ij})$, right?
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 15:47
$begingroup$
But you mean $Df(E_{ij})$, right?
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 15:47
$begingroup$
Actually, it's $Df_0$, that is, the derivative of $f$ at the null matrix. I've edited my answer.
$endgroup$
– José Carlos Santos
Jan 11 at 15:50
$begingroup$
Actually, it's $Df_0$, that is, the derivative of $f$ at the null matrix. I've edited my answer.
$endgroup$
– José Carlos Santos
Jan 11 at 15:50
$begingroup$
ah ok, I just realized that $Df(E_{ij})$ is as well not correct, because the $E_{ij}$ are not in G
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 15:52
$begingroup$
ah ok, I just realized that $Df(E_{ij})$ is as well not correct, because the $E_{ij}$ are not in G
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 15:52
$begingroup$
Ok thanks, but why to consider the derivative of $f$ at the null matrix? I don't understand it
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 16:00
$begingroup$
Ok thanks, but why to consider the derivative of $f$ at the null matrix? I don't understand it
$endgroup$
– TheAppliedTheoreticalOne
Jan 11 at 16:00
|
show 1 more comment
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