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Is the set of non invertible matrices simply connected? What are their homotopy and homology groups?
Two CW complexes with isomorphic homotopy groups and homology, yet not homotopy equivalentAlgebraic Topology pamphlets?If compact simply connected manifold has the same rational homotopy groups as $S^n$ or $mathbbCP^n$, must it have the same cohomology ring?Homology Whitehead theorem for non simply connected spacesCan we simultaneously realize arbitrary homotopy groups and arbitrary homology groups?Is there a non-simply-connected space with trivial first homology group?Connected odd dimensional Manifold with all its even homology groups are trivial except the zeroth homology group.Finiteness of homotopy groups and homology groups of H-spacesLens space and its generalization, homotopy/homology groupsHomology Groups of Oriented Simply Connected Manifold
$begingroup$
It is fairly easy to see that the set of non invertible matrices is path connected. Are they simply connected? If not what is their fundamental group? What are their homotopy and homology groups. I'm looking for the answer to any of these questions. Any examples for particular (non 0 or 1) dimensions are welcome as well, as well as for real or complex coefficients.
(Sorry about the phrasing, it is currently 3:30 am and I will edit the question in the morning)
linear-algebra algebraic-topology
edited May 2 at 20:56
Eric Wofsey
196k14226358
asked May 2 at 7:32
LiquidLiquid
32010
$endgroup$
add a comment |
$begingroup$
It is fairly easy to see that the set of non invertible matrices is path connected. Are they simply connected? If not what is their fundamental group? What are their homotopy and homology groups. I'm looking for the answer to any of these questions. Any examples for particular (non 0 or 1) dimensions are welcome as well, as well as for real or complex coefficients.
(Sorry about the phrasing, it is currently 3:30 am and I will edit the question in the morning)
linear-algebra algebraic-topology
edited May 2 at 20:56
Eric Wofsey
196k14226358
asked May 2 at 7:32
LiquidLiquid
32010
$endgroup$
1
$begingroup$
Matrices over what field or ring? And of what dimension?
$endgroup$
– Servaes
May 2 at 7:37
1
$begingroup$
With real or complex coefficients, and for any dimension you can show.
$endgroup$
– Liquid
May 2 at 7:40
1
$begingroup$
I'm mostly curious to know if they're simply connected, as my intuition is that the set looks like a bunch of lower dimensional vector spaces glued together at the origin
$endgroup$
– Liquid
May 2 at 7:42
add a comment |
$begingroup$
It is fairly easy to see that the set of non invertible matrices is path connected. Are they simply connected? If not what is their fundamental group? What are their homotopy and homology groups. I'm looking for the answer to any of these questions. Any examples for particular (non 0 or 1) dimensions are welcome as well, as well as for real or complex coefficients.
(Sorry about the phrasing, it is currently 3:30 am and I will edit the question in the morning)
linear-algebra algebraic-topology
edited May 2 at 20:56
Eric Wofsey
196k14226358
asked May 2 at 7:32
LiquidLiquid
32010
$endgroup$
It is fairly easy to see that the set of non invertible matrices is path connected. Are they simply connected? If not what is their fundamental group? What are their homotopy and homology groups. I'm looking for the answer to any of these questions. Any examples for particular (non 0 or 1) dimensions are welcome as well, as well as for real or complex coefficients.
(Sorry about the phrasing, it is currently 3:30 am and I will edit the question in the morning)
linear-algebra algebraic-topology
linear-algebra algebraic-topology
edited May 2 at 20:56
Eric Wofsey
196k14226358
asked May 2 at 7:32
LiquidLiquid
32010
edited May 2 at 20:56
Eric Wofsey
196k14226358
asked May 2 at 7:32
LiquidLiquid
32010
edited May 2 at 20:56
Eric Wofsey
196k14226358
edited May 2 at 20:56
Eric Wofsey
196k14226358
edited May 2 at 20:56
Eric Wofsey
196k14226358
196k14226358
asked May 2 at 7:32
LiquidLiquid
32010
asked May 2 at 7:32
LiquidLiquid
32010
asked May 2 at 7:32
LiquidLiquid
32010
32010
1
$begingroup$
Matrices over what field or ring? And of what dimension?
$endgroup$
– Servaes
May 2 at 7:37
1
$begingroup$
With real or complex coefficients, and for any dimension you can show.
$endgroup$
– Liquid
May 2 at 7:40
1
$begingroup$
I'm mostly curious to know if they're simply connected, as my intuition is that the set looks like a bunch of lower dimensional vector spaces glued together at the origin
$endgroup$
– Liquid
May 2 at 7:42
add a comment |
1
$begingroup$
Matrices over what field or ring? And of what dimension?
$endgroup$
– Servaes
May 2 at 7:37
1
$begingroup$
With real or complex coefficients, and for any dimension you can show.
$endgroup$
– Liquid
May 2 at 7:40
1
$begingroup$
I'm mostly curious to know if they're simply connected, as my intuition is that the set looks like a bunch of lower dimensional vector spaces glued together at the origin
$endgroup$
– Liquid
May 2 at 7:42
1
1
$begingroup$
Matrices over what field or ring? And of what dimension?
$endgroup$
– Servaes
May 2 at 7:37
$begingroup$
Matrices over what field or ring? And of what dimension?
$endgroup$
– Servaes
May 2 at 7:37
1
1
$begingroup$
With real or complex coefficients, and for any dimension you can show.
$endgroup$
– Liquid
May 2 at 7:40
$begingroup$
With real or complex coefficients, and for any dimension you can show.
$endgroup$
– Liquid
May 2 at 7:40
1
1
$begingroup$
I'm mostly curious to know if they're simply connected, as my intuition is that the set looks like a bunch of lower dimensional vector spaces glued together at the origin
$endgroup$
– Liquid
May 2 at 7:42
$begingroup$
I'm mostly curious to know if they're simply connected, as my intuition is that the set looks like a bunch of lower dimensional vector spaces glued together at the origin
$endgroup$
– Liquid
May 2 at 7:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The space of non-invertible matrices (with real or complex coefficients) is contractible. There is an explicit homotopy between the identity map and the constant map equal to the zero matrix, simply given by:
$$H(A,t) = tA.$$
In particular it is simply connected, and all its higher homotopy and homology groups vanish.
answered May 2 at 7:42
Najib IdrissiNajib Idrissi
42.6k476144
$endgroup$
2
$begingroup$
I didn't realize the space was star convex, I guess posting questions at 3am isn't the best idea.
$endgroup$
– Liquid
May 2 at 7:45
$begingroup$
@Liquid No big deal. When you learn all these sophisticated tools, the first impulse becomes to try using them, even though the answer is sometimes much simpler...
$endgroup$
– Najib Idrissi
May 2 at 7:46
$begingroup$
You are meaning the $0$ matrix and not the identity if I follow you well?
$endgroup$
– mathcounterexamples.net
May 2 at 7:47
1
$begingroup$
@mathcounterexamples.net I mean that the identity $H(A,1) = A$ is homotopic to the constant map equal to the zero matrix $H(A,0) = 0$.
$endgroup$
– Najib Idrissi
May 2 at 7:48
1
$begingroup$
It appears that we can conclude that either mathcounterexamples.net is not familiar with homotopy and contractibility, or your assertion is false.
$endgroup$
– Acccumulation
May 2 at 16:04
|
show 8 more comments
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1 Answer
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
The space of non-invertible matrices (with real or complex coefficients) is contractible. There is an explicit homotopy between the identity map and the constant map equal to the zero matrix, simply given by:
$$H(A,t) = tA.$$
In particular it is simply connected, and all its higher homotopy and homology groups vanish.
answered May 2 at 7:42
Najib IdrissiNajib Idrissi
42.6k476144
$endgroup$
2
$begingroup$
I didn't realize the space was star convex, I guess posting questions at 3am isn't the best idea.
$endgroup$
– Liquid
May 2 at 7:45
$begingroup$
@Liquid No big deal. When you learn all these sophisticated tools, the first impulse becomes to try using them, even though the answer is sometimes much simpler...
$endgroup$
– Najib Idrissi
May 2 at 7:46
$begingroup$
You are meaning the $0$ matrix and not the identity if I follow you well?
$endgroup$
– mathcounterexamples.net
May 2 at 7:47
1
$begingroup$
@mathcounterexamples.net I mean that the identity $H(A,1) = A$ is homotopic to the constant map equal to the zero matrix $H(A,0) = 0$.
$endgroup$
– Najib Idrissi
May 2 at 7:48
1
$begingroup$
It appears that we can conclude that either mathcounterexamples.net is not familiar with homotopy and contractibility, or your assertion is false.
$endgroup$
– Acccumulation
May 2 at 16:04
|
show 8 more comments
$begingroup$
The space of non-invertible matrices (with real or complex coefficients) is contractible. There is an explicit homotopy between the identity map and the constant map equal to the zero matrix, simply given by:
$$H(A,t) = tA.$$
In particular it is simply connected, and all its higher homotopy and homology groups vanish.
answered May 2 at 7:42
Najib IdrissiNajib Idrissi
42.6k476144
$endgroup$
2
$begingroup$
I didn't realize the space was star convex, I guess posting questions at 3am isn't the best idea.
$endgroup$
– Liquid
May 2 at 7:45
$begingroup$
@Liquid No big deal. When you learn all these sophisticated tools, the first impulse becomes to try using them, even though the answer is sometimes much simpler...
$endgroup$
– Najib Idrissi
May 2 at 7:46
$begingroup$
You are meaning the $0$ matrix and not the identity if I follow you well?
$endgroup$
– mathcounterexamples.net
May 2 at 7:47
1
$begingroup$
@mathcounterexamples.net I mean that the identity $H(A,1) = A$ is homotopic to the constant map equal to the zero matrix $H(A,0) = 0$.
$endgroup$
– Najib Idrissi
May 2 at 7:48
1
$begingroup$
It appears that we can conclude that either mathcounterexamples.net is not familiar with homotopy and contractibility, or your assertion is false.
$endgroup$
– Acccumulation
May 2 at 16:04
|
show 8 more comments
$begingroup$
The space of non-invertible matrices (with real or complex coefficients) is contractible. There is an explicit homotopy between the identity map and the constant map equal to the zero matrix, simply given by:
$$H(A,t) = tA.$$
In particular it is simply connected, and all its higher homotopy and homology groups vanish.
answered May 2 at 7:42
Najib IdrissiNajib Idrissi
42.6k476144
$endgroup$
The space of non-invertible matrices (with real or complex coefficients) is contractible. There is an explicit homotopy between the identity map and the constant map equal to the zero matrix, simply given by:
$$H(A,t) = tA.$$
In particular it is simply connected, and all its higher homotopy and homology groups vanish.
answered May 2 at 7:42
Najib IdrissiNajib Idrissi
42.6k476144
edited May 2 at 15:46
answered May 2 at 7:42
Najib IdrissiNajib Idrissi
42.6k476144
answered May 2 at 7:42
Najib IdrissiNajib Idrissi
42.6k476144
answered May 2 at 7:42
Najib IdrissiNajib Idrissi
42.6k476144
42.6k476144
2
$begingroup$
I didn't realize the space was star convex, I guess posting questions at 3am isn't the best idea.
$endgroup$
– Liquid
May 2 at 7:45
$begingroup$
@Liquid No big deal. When you learn all these sophisticated tools, the first impulse becomes to try using them, even though the answer is sometimes much simpler...
$endgroup$
– Najib Idrissi
May 2 at 7:46
$begingroup$
You are meaning the $0$ matrix and not the identity if I follow you well?
$endgroup$
– mathcounterexamples.net
May 2 at 7:47
1
$begingroup$
@mathcounterexamples.net I mean that the identity $H(A,1) = A$ is homotopic to the constant map equal to the zero matrix $H(A,0) = 0$.
$endgroup$
– Najib Idrissi
May 2 at 7:48
1
$begingroup$
It appears that we can conclude that either mathcounterexamples.net is not familiar with homotopy and contractibility, or your assertion is false.
$endgroup$
– Acccumulation
May 2 at 16:04
|
show 8 more comments
2
$begingroup$
I didn't realize the space was star convex, I guess posting questions at 3am isn't the best idea.
$endgroup$
– Liquid
May 2 at 7:45
$begingroup$
@Liquid No big deal. When you learn all these sophisticated tools, the first impulse becomes to try using them, even though the answer is sometimes much simpler...
$endgroup$
– Najib Idrissi
May 2 at 7:46
$begingroup$
You are meaning the $0$ matrix and not the identity if I follow you well?
$endgroup$
– mathcounterexamples.net
May 2 at 7:47
1
$begingroup$
@mathcounterexamples.net I mean that the identity $H(A,1) = A$ is homotopic to the constant map equal to the zero matrix $H(A,0) = 0$.
$endgroup$
– Najib Idrissi
May 2 at 7:48
1
$begingroup$
It appears that we can conclude that either mathcounterexamples.net is not familiar with homotopy and contractibility, or your assertion is false.
$endgroup$
– Acccumulation
May 2 at 16:04
2
2
$begingroup$
I didn't realize the space was star convex, I guess posting questions at 3am isn't the best idea.
$endgroup$
– Liquid
May 2 at 7:45
$begingroup$
I didn't realize the space was star convex, I guess posting questions at 3am isn't the best idea.
$endgroup$
– Liquid
May 2 at 7:45
$begingroup$
@Liquid No big deal. When you learn all these sophisticated tools, the first impulse becomes to try using them, even though the answer is sometimes much simpler...
$endgroup$
– Najib Idrissi
May 2 at 7:46
$begingroup$
@Liquid No big deal. When you learn all these sophisticated tools, the first impulse becomes to try using them, even though the answer is sometimes much simpler...
$endgroup$
– Najib Idrissi
May 2 at 7:46
$begingroup$
You are meaning the $0$ matrix and not the identity if I follow you well?
$endgroup$
– mathcounterexamples.net
May 2 at 7:47
$begingroup$
You are meaning the $0$ matrix and not the identity if I follow you well?
$endgroup$
– mathcounterexamples.net
May 2 at 7:47
1
1
$begingroup$
@mathcounterexamples.net I mean that the identity $H(A,1) = A$ is homotopic to the constant map equal to the zero matrix $H(A,0) = 0$.
$endgroup$
– Najib Idrissi
May 2 at 7:48
$begingroup$
@mathcounterexamples.net I mean that the identity $H(A,1) = A$ is homotopic to the constant map equal to the zero matrix $H(A,0) = 0$.
$endgroup$
– Najib Idrissi
May 2 at 7:48
1
1
$begingroup$
It appears that we can conclude that either mathcounterexamples.net is not familiar with homotopy and contractibility, or your assertion is false.
$endgroup$
– Acccumulation
May 2 at 16:04
$begingroup$
It appears that we can conclude that either mathcounterexamples.net is not familiar with homotopy and contractibility, or your assertion is false.
$endgroup$
– Acccumulation
May 2 at 16:04
|
show 8 more comments
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1
$begingroup$
Matrices over what field or ring? And of what dimension?
$endgroup$
– Servaes
May 2 at 7:37
1
$begingroup$
With real or complex coefficients, and for any dimension you can show.
$endgroup$
– Liquid
May 2 at 7:40
1
$begingroup$
I'm mostly curious to know if they're simply connected, as my intuition is that the set looks like a bunch of lower dimensional vector spaces glued together at the origin
$endgroup$
– Liquid
May 2 at 7:42