If V and W are linear subspaces of $R^n$, then V transversal W means just $V + W = R^n$
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If V and W are linear subspaces of $R^n$, then V transversal W means just $V + W = R^n$.
Could anyone give me a hint for this exercise please?
linear-algebra definition differential-topology geometric-topology transversality
asked Nov 21 at 17:04
hopefully
10312
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up vote
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If V and W are linear subspaces of $R^n$, then V transversal W means just $V + W = R^n$.
Could anyone give me a hint for this exercise please?
linear-algebra definition differential-topology geometric-topology transversality
asked Nov 21 at 17:04
hopefully
10312
2
What is your question? I don't think this is the usual definition for transversal.
– Ethan Bolker
Nov 21 at 17:05
2
This sounds more like a definition than something in need of proof.
– Rocket Man
Nov 21 at 17:06
What's up with all the random tags, but no "linear algebra"?
– Federico
Nov 21 at 17:10
This is an exercise in Guillemin & Pollack 1.5.1(b) @EthanBolker
– hopefully
Nov 21 at 17:14
@RocketMan This is an exercise in Guillemin & Pollack 1.5.1(b) - differential topology
– hopefully
Nov 21 at 17:16
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If V and W are linear subspaces of $R^n$, then V transversal W means just $V + W = R^n$.
Could anyone give me a hint for this exercise please?
linear-algebra definition differential-topology geometric-topology transversality
asked Nov 21 at 17:04
hopefully
10312
If V and W are linear subspaces of $R^n$, then V transversal W means just $V + W = R^n$.
Could anyone give me a hint for this exercise please?
linear-algebra definition differential-topology geometric-topology transversality
linear-algebra definition differential-topology geometric-topology transversality
asked Nov 21 at 17:04
hopefully
10312
asked Nov 21 at 17:04
hopefully
10312
edited Nov 21 at 17:15
asked Nov 21 at 17:04
hopefully
10312
asked Nov 21 at 17:04
hopefully
10312
asked Nov 21 at 17:04
hopefully
10312
10312
2
What is your question? I don't think this is the usual definition for transversal.
– Ethan Bolker
Nov 21 at 17:05
2
This sounds more like a definition than something in need of proof.
– Rocket Man
Nov 21 at 17:06
What's up with all the random tags, but no "linear algebra"?
– Federico
Nov 21 at 17:10
This is an exercise in Guillemin & Pollack 1.5.1(b) @EthanBolker
– hopefully
Nov 21 at 17:14
@RocketMan This is an exercise in Guillemin & Pollack 1.5.1(b) - differential topology
– hopefully
Nov 21 at 17:16
|
show 1 more comment
2
What is your question? I don't think this is the usual definition for transversal.
– Ethan Bolker
Nov 21 at 17:05
2
This sounds more like a definition than something in need of proof.
– Rocket Man
Nov 21 at 17:06
What's up with all the random tags, but no "linear algebra"?
– Federico
Nov 21 at 17:10
This is an exercise in Guillemin & Pollack 1.5.1(b) @EthanBolker
– hopefully
Nov 21 at 17:14
@RocketMan This is an exercise in Guillemin & Pollack 1.5.1(b) - differential topology
– hopefully
Nov 21 at 17:16
2
2
What is your question? I don't think this is the usual definition for transversal.
– Ethan Bolker
Nov 21 at 17:05
What is your question? I don't think this is the usual definition for transversal.
– Ethan Bolker
Nov 21 at 17:05
2
2
This sounds more like a definition than something in need of proof.
– Rocket Man
Nov 21 at 17:06
This sounds more like a definition than something in need of proof.
– Rocket Man
Nov 21 at 17:06
What's up with all the random tags, but no "linear algebra"?
– Federico
Nov 21 at 17:10
What's up with all the random tags, but no "linear algebra"?
– Federico
Nov 21 at 17:10
This is an exercise in Guillemin & Pollack 1.5.1(b) @EthanBolker
– hopefully
Nov 21 at 17:14
This is an exercise in Guillemin & Pollack 1.5.1(b) @EthanBolker
– hopefully
Nov 21 at 17:14
@RocketMan This is an exercise in Guillemin & Pollack 1.5.1(b) - differential topology
– hopefully
Nov 21 at 17:16
@RocketMan This is an exercise in Guillemin & Pollack 1.5.1(b) - differential topology
– hopefully
Nov 21 at 17:16
|
show 1 more comment
1 Answer
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So to answer this question, you have to think of $V$ and $W$ as submanifolds of $mathbb{R}^n$. They intersect transversally if, at every point of their intersection, the sum of their tangent spaces is $mathbb{R}^n$. But remember that since $V$ and $W$ are vector spaces, their tangent spaces at any point are isomorphically just themselves, and we identify them as such. The same goes for the tangent space to $mathbb{R}^n$ at any point. Since all these identifications commute in a sense that can be made precise, there is no conflict if we make all these identifications simultaneously and then think about just the subspaces themselves. Thinking about what this means in the context of transversality, we realize that this is just the condition that $V+W = mathbb{R}^n$.
answered Nov 21 at 17:32
Monstrous Moonshiner
2,25911337
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
So to answer this question, you have to think of $V$ and $W$ as submanifolds of $mathbb{R}^n$. They intersect transversally if, at every point of their intersection, the sum of their tangent spaces is $mathbb{R}^n$. But remember that since $V$ and $W$ are vector spaces, their tangent spaces at any point are isomorphically just themselves, and we identify them as such. The same goes for the tangent space to $mathbb{R}^n$ at any point. Since all these identifications commute in a sense that can be made precise, there is no conflict if we make all these identifications simultaneously and then think about just the subspaces themselves. Thinking about what this means in the context of transversality, we realize that this is just the condition that $V+W = mathbb{R}^n$.
answered Nov 21 at 17:32
Monstrous Moonshiner
2,25911337
add a comment |
up vote
1
down vote
accepted
So to answer this question, you have to think of $V$ and $W$ as submanifolds of $mathbb{R}^n$. They intersect transversally if, at every point of their intersection, the sum of their tangent spaces is $mathbb{R}^n$. But remember that since $V$ and $W$ are vector spaces, their tangent spaces at any point are isomorphically just themselves, and we identify them as such. The same goes for the tangent space to $mathbb{R}^n$ at any point. Since all these identifications commute in a sense that can be made precise, there is no conflict if we make all these identifications simultaneously and then think about just the subspaces themselves. Thinking about what this means in the context of transversality, we realize that this is just the condition that $V+W = mathbb{R}^n$.
answered Nov 21 at 17:32
Monstrous Moonshiner
2,25911337
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
So to answer this question, you have to think of $V$ and $W$ as submanifolds of $mathbb{R}^n$. They intersect transversally if, at every point of their intersection, the sum of their tangent spaces is $mathbb{R}^n$. But remember that since $V$ and $W$ are vector spaces, their tangent spaces at any point are isomorphically just themselves, and we identify them as such. The same goes for the tangent space to $mathbb{R}^n$ at any point. Since all these identifications commute in a sense that can be made precise, there is no conflict if we make all these identifications simultaneously and then think about just the subspaces themselves. Thinking about what this means in the context of transversality, we realize that this is just the condition that $V+W = mathbb{R}^n$.
answered Nov 21 at 17:32
Monstrous Moonshiner
2,25911337
So to answer this question, you have to think of $V$ and $W$ as submanifolds of $mathbb{R}^n$. They intersect transversally if, at every point of their intersection, the sum of their tangent spaces is $mathbb{R}^n$. But remember that since $V$ and $W$ are vector spaces, their tangent spaces at any point are isomorphically just themselves, and we identify them as such. The same goes for the tangent space to $mathbb{R}^n$ at any point. Since all these identifications commute in a sense that can be made precise, there is no conflict if we make all these identifications simultaneously and then think about just the subspaces themselves. Thinking about what this means in the context of transversality, we realize that this is just the condition that $V+W = mathbb{R}^n$.
answered Nov 21 at 17:32
Monstrous Moonshiner
2,25911337
answered Nov 21 at 17:32
Monstrous Moonshiner
2,25911337
answered Nov 21 at 17:32
Monstrous Moonshiner
2,25911337
answered Nov 21 at 17:32
Monstrous Moonshiner
2,25911337
2,25911337
add a comment |
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2
What is your question? I don't think this is the usual definition for transversal.
– Ethan Bolker
Nov 21 at 17:05
2
This sounds more like a definition than something in need of proof.
– Rocket Man
Nov 21 at 17:06
What's up with all the random tags, but no "linear algebra"?
– Federico
Nov 21 at 17:10
This is an exercise in Guillemin & Pollack 1.5.1(b) @EthanBolker
– hopefully
Nov 21 at 17:14
@RocketMan This is an exercise in Guillemin & Pollack 1.5.1(b) - differential topology
– hopefully
Nov 21 at 17:16