Why is the identity component of a matrix group a subgroup?
$begingroup$
I'm working through Stillwell's "Naive Lie Theory". I'm supposed to show that the identity component of a matrix group is a subgroup in two steps. I'm allowed to assume that "matrix multiplication is a continuous operation". First question- what does this mean? Does this mean multiplying matrices by a fixed matrix is continuous, or multiplying two matrices which vary?
In the first step, I'm supposed to prove that if there are continuous paths in the group $G$ from 1 to $A in G$ and to $B in G$ then there is a path in G from $A$ to $AB$.
I did this by assuming that matrix multiplication by a fixed matrix was continuous. I presume that this will get us closure under group operation by concatenating the path from 1 to $A$ with the path from $A$ to $AB$.
Second, and where I am stuck, is in proving that if there is a continuous path in $G$ from 1 to $A$ there is also a continuous path from $A^{-1}$ to 1. If I knew that the map that sends $A$ to $A^{-1}$ was continuous, I think I would be done, but I don't know how to get this easily.
matrices lie-groups
asked May 11 '11 at 1:20
Aron SamkoffAron Samkoff
161
$endgroup$
add a comment |
$begingroup$
I'm working through Stillwell's "Naive Lie Theory". I'm supposed to show that the identity component of a matrix group is a subgroup in two steps. I'm allowed to assume that "matrix multiplication is a continuous operation". First question- what does this mean? Does this mean multiplying matrices by a fixed matrix is continuous, or multiplying two matrices which vary?
In the first step, I'm supposed to prove that if there are continuous paths in the group $G$ from 1 to $A in G$ and to $B in G$ then there is a path in G from $A$ to $AB$.
I did this by assuming that matrix multiplication by a fixed matrix was continuous. I presume that this will get us closure under group operation by concatenating the path from 1 to $A$ with the path from $A$ to $AB$.
Second, and where I am stuck, is in proving that if there is a continuous path in $G$ from 1 to $A$ there is also a continuous path from $A^{-1}$ to 1. If I knew that the map that sends $A$ to $A^{-1}$ was continuous, I think I would be done, but I don't know how to get this easily.
matrices lie-groups
asked May 11 '11 at 1:20
Aron SamkoffAron Samkoff
161
$endgroup$
$begingroup$
Just a comment: it is not hard to show that the connected component of the identity of any topological group is a normal subgroup. It sounds like Stillwell is taking a more "naive" perspective than this, but this much naivete may not be to everyone's taste...
$endgroup$
– Pete L. Clark
May 11 '11 at 2:17
add a comment |
$begingroup$
I'm working through Stillwell's "Naive Lie Theory". I'm supposed to show that the identity component of a matrix group is a subgroup in two steps. I'm allowed to assume that "matrix multiplication is a continuous operation". First question- what does this mean? Does this mean multiplying matrices by a fixed matrix is continuous, or multiplying two matrices which vary?
In the first step, I'm supposed to prove that if there are continuous paths in the group $G$ from 1 to $A in G$ and to $B in G$ then there is a path in G from $A$ to $AB$.
I did this by assuming that matrix multiplication by a fixed matrix was continuous. I presume that this will get us closure under group operation by concatenating the path from 1 to $A$ with the path from $A$ to $AB$.
Second, and where I am stuck, is in proving that if there is a continuous path in $G$ from 1 to $A$ there is also a continuous path from $A^{-1}$ to 1. If I knew that the map that sends $A$ to $A^{-1}$ was continuous, I think I would be done, but I don't know how to get this easily.
matrices lie-groups
asked May 11 '11 at 1:20
Aron SamkoffAron Samkoff
161
$endgroup$
I'm working through Stillwell's "Naive Lie Theory". I'm supposed to show that the identity component of a matrix group is a subgroup in two steps. I'm allowed to assume that "matrix multiplication is a continuous operation". First question- what does this mean? Does this mean multiplying matrices by a fixed matrix is continuous, or multiplying two matrices which vary?
In the first step, I'm supposed to prove that if there are continuous paths in the group $G$ from 1 to $A in G$ and to $B in G$ then there is a path in G from $A$ to $AB$.
I did this by assuming that matrix multiplication by a fixed matrix was continuous. I presume that this will get us closure under group operation by concatenating the path from 1 to $A$ with the path from $A$ to $AB$.
Second, and where I am stuck, is in proving that if there is a continuous path in $G$ from 1 to $A$ there is also a continuous path from $A^{-1}$ to 1. If I knew that the map that sends $A$ to $A^{-1}$ was continuous, I think I would be done, but I don't know how to get this easily.
matrices lie-groups
matrices lie-groups
asked May 11 '11 at 1:20
Aron SamkoffAron Samkoff
161
asked May 11 '11 at 1:20
Aron SamkoffAron Samkoff
161
asked May 11 '11 at 1:20
Aron SamkoffAron Samkoff
161
asked May 11 '11 at 1:20
Aron SamkoffAron Samkoff
161
asked May 11 '11 at 1:20
Aron SamkoffAron Samkoff
161
161
$begingroup$
Just a comment: it is not hard to show that the connected component of the identity of any topological group is a normal subgroup. It sounds like Stillwell is taking a more "naive" perspective than this, but this much naivete may not be to everyone's taste...
$endgroup$
– Pete L. Clark
May 11 '11 at 2:17
add a comment |
$begingroup$
Just a comment: it is not hard to show that the connected component of the identity of any topological group is a normal subgroup. It sounds like Stillwell is taking a more "naive" perspective than this, but this much naivete may not be to everyone's taste...
$endgroup$
– Pete L. Clark
May 11 '11 at 2:17
$begingroup$
Just a comment: it is not hard to show that the connected component of the identity of any topological group is a normal subgroup. It sounds like Stillwell is taking a more "naive" perspective than this, but this much naivete may not be to everyone's taste...
$endgroup$
– Pete L. Clark
May 11 '11 at 2:17
$begingroup$
Just a comment: it is not hard to show that the connected component of the identity of any topological group is a normal subgroup. It sounds like Stillwell is taking a more "naive" perspective than this, but this much naivete may not be to everyone's taste...
$endgroup$
– Pete L. Clark
May 11 '11 at 2:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Multiplying two matrices. That is, $times : G times G to G$ is continuous.
You don't need to know that inversion is continuous (although it is by Cramer's rule). You just need to multiply the path by $A^{-1}$.
answered May 11 '11 at 1:30
Qiaochu YuanQiaochu Yuan
278k32585921
$endgroup$
$begingroup$
Well, that was easy :) Thanks. Then can I ask, why is this function that takes pairs of matrices to their product continuous?
$endgroup$
– Aron Samkoff
May 11 '11 at 1:35
2
$begingroup$
@Aron: it suffices to show that multiplying matrices by matrices is continuous, and for that it suffices to observe that the result of multiplying two matrices is a matrix whose entries are polynomials in the entries of the matrices, and polynomials are always continuous.
$endgroup$
– Qiaochu Yuan
May 11 '11 at 1:37
add a comment |
$begingroup$
Hint. The path component and the connected component of this group containing the identity are one in the same.
answered May 11 '11 at 1:29
ncmathsadistncmathsadist
42.7k260103
$endgroup$
1
$begingroup$
That isn't the issue the OP is having.
$endgroup$
– Qiaochu Yuan
May 11 '11 at 1:30
$begingroup$
A standard method ( it is a theorem, actually) for showing that s subset S of a group is a subgroup of G is showing that if a,b are in S, then the product a$b^-1$ is also in S. You can then also use the fact that continuous maps preserve connected components, i.e., if we have a continuous map f:X-->Y , and X has components $C_1$,..,$C_n$, then f(X) will have also exactly n components f($C_1$),..,f($C_n$). By continuity, then, the component of the identity will be sent to itself under the map A$B^-1$, showing that the component of the identity is a subgroup.
$endgroup$
– gary
May 11 '11 at 1:56
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Multiplying two matrices. That is, $times : G times G to G$ is continuous.
You don't need to know that inversion is continuous (although it is by Cramer's rule). You just need to multiply the path by $A^{-1}$.
answered May 11 '11 at 1:30
Qiaochu YuanQiaochu Yuan
278k32585921
$endgroup$
$begingroup$
Well, that was easy :) Thanks. Then can I ask, why is this function that takes pairs of matrices to their product continuous?
$endgroup$
– Aron Samkoff
May 11 '11 at 1:35
2
$begingroup$
@Aron: it suffices to show that multiplying matrices by matrices is continuous, and for that it suffices to observe that the result of multiplying two matrices is a matrix whose entries are polynomials in the entries of the matrices, and polynomials are always continuous.
$endgroup$
– Qiaochu Yuan
May 11 '11 at 1:37
add a comment |
$begingroup$
Multiplying two matrices. That is, $times : G times G to G$ is continuous.
You don't need to know that inversion is continuous (although it is by Cramer's rule). You just need to multiply the path by $A^{-1}$.
answered May 11 '11 at 1:30
Qiaochu YuanQiaochu Yuan
278k32585921
$endgroup$
$begingroup$
Well, that was easy :) Thanks. Then can I ask, why is this function that takes pairs of matrices to their product continuous?
$endgroup$
– Aron Samkoff
May 11 '11 at 1:35
2
$begingroup$
@Aron: it suffices to show that multiplying matrices by matrices is continuous, and for that it suffices to observe that the result of multiplying two matrices is a matrix whose entries are polynomials in the entries of the matrices, and polynomials are always continuous.
$endgroup$
– Qiaochu Yuan
May 11 '11 at 1:37
add a comment |
$begingroup$
Multiplying two matrices. That is, $times : G times G to G$ is continuous.
You don't need to know that inversion is continuous (although it is by Cramer's rule). You just need to multiply the path by $A^{-1}$.
answered May 11 '11 at 1:30
Qiaochu YuanQiaochu Yuan
278k32585921
$endgroup$
Multiplying two matrices. That is, $times : G times G to G$ is continuous.
You don't need to know that inversion is continuous (although it is by Cramer's rule). You just need to multiply the path by $A^{-1}$.
answered May 11 '11 at 1:30
Qiaochu YuanQiaochu Yuan
278k32585921
edited May 11 '11 at 1:37
answered May 11 '11 at 1:30
Qiaochu YuanQiaochu Yuan
278k32585921
answered May 11 '11 at 1:30
Qiaochu YuanQiaochu Yuan
278k32585921
answered May 11 '11 at 1:30
Qiaochu YuanQiaochu Yuan
278k32585921
278k32585921
$begingroup$
Well, that was easy :) Thanks. Then can I ask, why is this function that takes pairs of matrices to their product continuous?
$endgroup$
– Aron Samkoff
May 11 '11 at 1:35
2
$begingroup$
@Aron: it suffices to show that multiplying matrices by matrices is continuous, and for that it suffices to observe that the result of multiplying two matrices is a matrix whose entries are polynomials in the entries of the matrices, and polynomials are always continuous.
$endgroup$
– Qiaochu Yuan
May 11 '11 at 1:37
add a comment |
$begingroup$
Well, that was easy :) Thanks. Then can I ask, why is this function that takes pairs of matrices to their product continuous?
$endgroup$
– Aron Samkoff
May 11 '11 at 1:35
2
$begingroup$
@Aron: it suffices to show that multiplying matrices by matrices is continuous, and for that it suffices to observe that the result of multiplying two matrices is a matrix whose entries are polynomials in the entries of the matrices, and polynomials are always continuous.
$endgroup$
– Qiaochu Yuan
May 11 '11 at 1:37
$begingroup$
Well, that was easy :) Thanks. Then can I ask, why is this function that takes pairs of matrices to their product continuous?
$endgroup$
– Aron Samkoff
May 11 '11 at 1:35
$begingroup$
Well, that was easy :) Thanks. Then can I ask, why is this function that takes pairs of matrices to their product continuous?
$endgroup$
– Aron Samkoff
May 11 '11 at 1:35
2
2
$begingroup$
@Aron: it suffices to show that multiplying matrices by matrices is continuous, and for that it suffices to observe that the result of multiplying two matrices is a matrix whose entries are polynomials in the entries of the matrices, and polynomials are always continuous.
$endgroup$
– Qiaochu Yuan
May 11 '11 at 1:37
$begingroup$
@Aron: it suffices to show that multiplying matrices by matrices is continuous, and for that it suffices to observe that the result of multiplying two matrices is a matrix whose entries are polynomials in the entries of the matrices, and polynomials are always continuous.
$endgroup$
– Qiaochu Yuan
May 11 '11 at 1:37
add a comment |
$begingroup$
Hint. The path component and the connected component of this group containing the identity are one in the same.
answered May 11 '11 at 1:29
ncmathsadistncmathsadist
42.7k260103
$endgroup$
1
$begingroup$
That isn't the issue the OP is having.
$endgroup$
– Qiaochu Yuan
May 11 '11 at 1:30
$begingroup$
A standard method ( it is a theorem, actually) for showing that s subset S of a group is a subgroup of G is showing that if a,b are in S, then the product a$b^-1$ is also in S. You can then also use the fact that continuous maps preserve connected components, i.e., if we have a continuous map f:X-->Y , and X has components $C_1$,..,$C_n$, then f(X) will have also exactly n components f($C_1$),..,f($C_n$). By continuity, then, the component of the identity will be sent to itself under the map A$B^-1$, showing that the component of the identity is a subgroup.
$endgroup$
– gary
May 11 '11 at 1:56
add a comment |
$begingroup$
Hint. The path component and the connected component of this group containing the identity are one in the same.
answered May 11 '11 at 1:29
ncmathsadistncmathsadist
42.7k260103
$endgroup$
1
$begingroup$
That isn't the issue the OP is having.
$endgroup$
– Qiaochu Yuan
May 11 '11 at 1:30
$begingroup$
A standard method ( it is a theorem, actually) for showing that s subset S of a group is a subgroup of G is showing that if a,b are in S, then the product a$b^-1$ is also in S. You can then also use the fact that continuous maps preserve connected components, i.e., if we have a continuous map f:X-->Y , and X has components $C_1$,..,$C_n$, then f(X) will have also exactly n components f($C_1$),..,f($C_n$). By continuity, then, the component of the identity will be sent to itself under the map A$B^-1$, showing that the component of the identity is a subgroup.
$endgroup$
– gary
May 11 '11 at 1:56
add a comment |
$begingroup$
Hint. The path component and the connected component of this group containing the identity are one in the same.
answered May 11 '11 at 1:29
ncmathsadistncmathsadist
42.7k260103
$endgroup$
Hint. The path component and the connected component of this group containing the identity are one in the same.
answered May 11 '11 at 1:29
ncmathsadistncmathsadist
42.7k260103
answered May 11 '11 at 1:29
ncmathsadistncmathsadist
42.7k260103
answered May 11 '11 at 1:29
ncmathsadistncmathsadist
42.7k260103
answered May 11 '11 at 1:29
ncmathsadistncmathsadist
42.7k260103
42.7k260103
1
$begingroup$
That isn't the issue the OP is having.
$endgroup$
– Qiaochu Yuan
May 11 '11 at 1:30
$begingroup$
A standard method ( it is a theorem, actually) for showing that s subset S of a group is a subgroup of G is showing that if a,b are in S, then the product a$b^-1$ is also in S. You can then also use the fact that continuous maps preserve connected components, i.e., if we have a continuous map f:X-->Y , and X has components $C_1$,..,$C_n$, then f(X) will have also exactly n components f($C_1$),..,f($C_n$). By continuity, then, the component of the identity will be sent to itself under the map A$B^-1$, showing that the component of the identity is a subgroup.
$endgroup$
– gary
May 11 '11 at 1:56
add a comment |
1
$begingroup$
That isn't the issue the OP is having.
$endgroup$
– Qiaochu Yuan
May 11 '11 at 1:30
$begingroup$
A standard method ( it is a theorem, actually) for showing that s subset S of a group is a subgroup of G is showing that if a,b are in S, then the product a$b^-1$ is also in S. You can then also use the fact that continuous maps preserve connected components, i.e., if we have a continuous map f:X-->Y , and X has components $C_1$,..,$C_n$, then f(X) will have also exactly n components f($C_1$),..,f($C_n$). By continuity, then, the component of the identity will be sent to itself under the map A$B^-1$, showing that the component of the identity is a subgroup.
$endgroup$
– gary
May 11 '11 at 1:56
1
1
$begingroup$
That isn't the issue the OP is having.
$endgroup$
– Qiaochu Yuan
May 11 '11 at 1:30
$begingroup$
That isn't the issue the OP is having.
$endgroup$
– Qiaochu Yuan
May 11 '11 at 1:30
$begingroup$
A standard method ( it is a theorem, actually) for showing that s subset S of a group is a subgroup of G is showing that if a,b are in S, then the product a$b^-1$ is also in S. You can then also use the fact that continuous maps preserve connected components, i.e., if we have a continuous map f:X-->Y , and X has components $C_1$,..,$C_n$, then f(X) will have also exactly n components f($C_1$),..,f($C_n$). By continuity, then, the component of the identity will be sent to itself under the map A$B^-1$, showing that the component of the identity is a subgroup.
$endgroup$
– gary
May 11 '11 at 1:56
$begingroup$
A standard method ( it is a theorem, actually) for showing that s subset S of a group is a subgroup of G is showing that if a,b are in S, then the product a$b^-1$ is also in S. You can then also use the fact that continuous maps preserve connected components, i.e., if we have a continuous map f:X-->Y , and X has components $C_1$,..,$C_n$, then f(X) will have also exactly n components f($C_1$),..,f($C_n$). By continuity, then, the component of the identity will be sent to itself under the map A$B^-1$, showing that the component of the identity is a subgroup.
$endgroup$
– gary
May 11 '11 at 1:56
add a comment |
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$begingroup$
Just a comment: it is not hard to show that the connected component of the identity of any topological group is a normal subgroup. It sounds like Stillwell is taking a more "naive" perspective than this, but this much naivete may not be to everyone's taste...
$endgroup$
– Pete L. Clark
May 11 '11 at 2:17