Embedding the quotient of a vector space
$begingroup$
Q: Given a vector space $V$ over an arbitrary field and a linear subspace $W$, can we embed $V/W$ within $V$?
An embedding would be an injective morphism $i: V/W to V $ which is an isomorphism from the domain to its image?
Also are there similar results on groups and embeddings of the quotients by their normal subgroups?
In the case of finite-dimensional real or complex vector spaces, we can use orthogonality, but I am unsure how this would work otherwise.
Related question: Is quotient space a subspace of the underlying vector space
linear-algebra abstract-algebra vector-spaces
asked Oct 25 '18 at 14:13
Marko KarbevskiMarko Karbevski
1,037821
$endgroup$
add a comment |
$begingroup$
Q: Given a vector space $V$ over an arbitrary field and a linear subspace $W$, can we embed $V/W$ within $V$?
An embedding would be an injective morphism $i: V/W to V $ which is an isomorphism from the domain to its image?
Also are there similar results on groups and embeddings of the quotients by their normal subgroups?
In the case of finite-dimensional real or complex vector spaces, we can use orthogonality, but I am unsure how this would work otherwise.
Related question: Is quotient space a subspace of the underlying vector space
linear-algebra abstract-algebra vector-spaces
asked Oct 25 '18 at 14:13
Marko KarbevskiMarko Karbevski
1,037821
$endgroup$
$begingroup$
The last assertion is false: For general finite-dimensional real or complex vector spaces there is no notion of orthogonality---this requires additional data (e.g., an inner product or Hermitian product).
$endgroup$
– Travis
Oct 26 '18 at 13:00
add a comment |
$begingroup$
Q: Given a vector space $V$ over an arbitrary field and a linear subspace $W$, can we embed $V/W$ within $V$?
An embedding would be an injective morphism $i: V/W to V $ which is an isomorphism from the domain to its image?
Also are there similar results on groups and embeddings of the quotients by their normal subgroups?
In the case of finite-dimensional real or complex vector spaces, we can use orthogonality, but I am unsure how this would work otherwise.
Related question: Is quotient space a subspace of the underlying vector space
linear-algebra abstract-algebra vector-spaces
asked Oct 25 '18 at 14:13
Marko KarbevskiMarko Karbevski
1,037821
$endgroup$
Q: Given a vector space $V$ over an arbitrary field and a linear subspace $W$, can we embed $V/W$ within $V$?
An embedding would be an injective morphism $i: V/W to V $ which is an isomorphism from the domain to its image?
Also are there similar results on groups and embeddings of the quotients by their normal subgroups?
In the case of finite-dimensional real or complex vector spaces, we can use orthogonality, but I am unsure how this would work otherwise.
Related question: Is quotient space a subspace of the underlying vector space
linear-algebra abstract-algebra vector-spaces
linear-algebra abstract-algebra vector-spaces
asked Oct 25 '18 at 14:13
Marko KarbevskiMarko Karbevski
1,037821
asked Oct 25 '18 at 14:13
Marko KarbevskiMarko Karbevski
1,037821
asked Oct 25 '18 at 14:13
Marko KarbevskiMarko Karbevski
1,037821
asked Oct 25 '18 at 14:13
Marko KarbevskiMarko Karbevski
1,037821
asked Oct 25 '18 at 14:13
Marko KarbevskiMarko Karbevski
1,037821
1,037821
$begingroup$
The last assertion is false: For general finite-dimensional real or complex vector spaces there is no notion of orthogonality---this requires additional data (e.g., an inner product or Hermitian product).
$endgroup$
– Travis
Oct 26 '18 at 13:00
add a comment |
$begingroup$
The last assertion is false: For general finite-dimensional real or complex vector spaces there is no notion of orthogonality---this requires additional data (e.g., an inner product or Hermitian product).
$endgroup$
– Travis
Oct 26 '18 at 13:00
$begingroup$
The last assertion is false: For general finite-dimensional real or complex vector spaces there is no notion of orthogonality---this requires additional data (e.g., an inner product or Hermitian product).
$endgroup$
– Travis
Oct 26 '18 at 13:00
$begingroup$
The last assertion is false: For general finite-dimensional real or complex vector spaces there is no notion of orthogonality---this requires additional data (e.g., an inner product or Hermitian product).
$endgroup$
– Travis
Oct 26 '18 at 13:00
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
SO, let's back up. For groups or rings, there is no general guarantee that we can imbed a quotient into the group/ring itself. This is true exactly when the kernel has a compliment in the group/ring itself.
Vector spaces are special though. They can always be written as a direct sum of their one dimensional subspaces (with respect to a given ordered basis). So, in this case where we have (I will only demonstrate the finite dim. case but it is similar for infinite dim. at least with countable bases)
$$V=V_1 oplus V_2 oplus dots oplus V_n$$
So, if we quotient out by a particular subpsace then we are just quotienting at by some ``subsum'' of the above essentially. That is, the kernel of that map (the thing you're quotienting out by) you have always has a compliment. That compliment is how we imbed $V/K$ into $V$. It takes very little work to show $V cong V oplus V/K$
answered Oct 25 '18 at 14:19
RhythmInkRhythmInk
1,683423
$endgroup$
add a comment |
$begingroup$
Yes. Take a basis ${e_i}$ of $V/W$, and choose $f_i$ in $V$ such that $f_i$ maps to $e_i$ under the quotient map. The map from $V/W$ to $V$ linearly extending the above gives an embedding. If $sum a_i e_i = z in V/W$ (a finite sum) then the image in $V$ would be $sum a_i f_i$. etc
edited Oct 25 '18 at 17:13
memerson
55510
answered Oct 25 '18 at 16:31
R.C.CowsikR.C.Cowsik
31514
$endgroup$
add a comment |
$begingroup$
I completely agree with the previous answers. However, please be aware that any such an embedding is actually a choice! So there is no canonical way on doing it (it is only canonical if you quotient out a trivial space, i.e. the whole space or 0). Furthermore, if you are interested, this property is usually referred to as semi-simplicity of a category (something quite powerful).
and to get back to Rhythmink's answer: an example for rings or groups were this is not possible is the map $$mathbb{Z} to mathbb{Z}/(p).$$
I leave it to you to show that this admits no leftinverse ,i.e. an embedding
$$mathbb{Z}/(p) to mathbb{Z}.$$
answered Oct 26 '18 at 8:42
EnkiduEnkidu
1,32619
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
SO, let's back up. For groups or rings, there is no general guarantee that we can imbed a quotient into the group/ring itself. This is true exactly when the kernel has a compliment in the group/ring itself.
Vector spaces are special though. They can always be written as a direct sum of their one dimensional subspaces (with respect to a given ordered basis). So, in this case where we have (I will only demonstrate the finite dim. case but it is similar for infinite dim. at least with countable bases)
$$V=V_1 oplus V_2 oplus dots oplus V_n$$
So, if we quotient out by a particular subpsace then we are just quotienting at by some ``subsum'' of the above essentially. That is, the kernel of that map (the thing you're quotienting out by) you have always has a compliment. That compliment is how we imbed $V/K$ into $V$. It takes very little work to show $V cong V oplus V/K$
answered Oct 25 '18 at 14:19
RhythmInkRhythmInk
1,683423
$endgroup$
add a comment |
$begingroup$
SO, let's back up. For groups or rings, there is no general guarantee that we can imbed a quotient into the group/ring itself. This is true exactly when the kernel has a compliment in the group/ring itself.
Vector spaces are special though. They can always be written as a direct sum of their one dimensional subspaces (with respect to a given ordered basis). So, in this case where we have (I will only demonstrate the finite dim. case but it is similar for infinite dim. at least with countable bases)
$$V=V_1 oplus V_2 oplus dots oplus V_n$$
So, if we quotient out by a particular subpsace then we are just quotienting at by some ``subsum'' of the above essentially. That is, the kernel of that map (the thing you're quotienting out by) you have always has a compliment. That compliment is how we imbed $V/K$ into $V$. It takes very little work to show $V cong V oplus V/K$
answered Oct 25 '18 at 14:19
RhythmInkRhythmInk
1,683423
$endgroup$
add a comment |
$begingroup$
SO, let's back up. For groups or rings, there is no general guarantee that we can imbed a quotient into the group/ring itself. This is true exactly when the kernel has a compliment in the group/ring itself.
Vector spaces are special though. They can always be written as a direct sum of their one dimensional subspaces (with respect to a given ordered basis). So, in this case where we have (I will only demonstrate the finite dim. case but it is similar for infinite dim. at least with countable bases)
$$V=V_1 oplus V_2 oplus dots oplus V_n$$
So, if we quotient out by a particular subpsace then we are just quotienting at by some ``subsum'' of the above essentially. That is, the kernel of that map (the thing you're quotienting out by) you have always has a compliment. That compliment is how we imbed $V/K$ into $V$. It takes very little work to show $V cong V oplus V/K$
answered Oct 25 '18 at 14:19
RhythmInkRhythmInk
1,683423
$endgroup$
SO, let's back up. For groups or rings, there is no general guarantee that we can imbed a quotient into the group/ring itself. This is true exactly when the kernel has a compliment in the group/ring itself.
Vector spaces are special though. They can always be written as a direct sum of their one dimensional subspaces (with respect to a given ordered basis). So, in this case where we have (I will only demonstrate the finite dim. case but it is similar for infinite dim. at least with countable bases)
$$V=V_1 oplus V_2 oplus dots oplus V_n$$
So, if we quotient out by a particular subpsace then we are just quotienting at by some ``subsum'' of the above essentially. That is, the kernel of that map (the thing you're quotienting out by) you have always has a compliment. That compliment is how we imbed $V/K$ into $V$. It takes very little work to show $V cong V oplus V/K$
answered Oct 25 '18 at 14:19
RhythmInkRhythmInk
1,683423
answered Oct 25 '18 at 14:19
RhythmInkRhythmInk
1,683423
answered Oct 25 '18 at 14:19
RhythmInkRhythmInk
1,683423
answered Oct 25 '18 at 14:19
RhythmInkRhythmInk
1,683423
1,683423
add a comment |
add a comment |
$begingroup$
Yes. Take a basis ${e_i}$ of $V/W$, and choose $f_i$ in $V$ such that $f_i$ maps to $e_i$ under the quotient map. The map from $V/W$ to $V$ linearly extending the above gives an embedding. If $sum a_i e_i = z in V/W$ (a finite sum) then the image in $V$ would be $sum a_i f_i$. etc
edited Oct 25 '18 at 17:13
memerson
55510
answered Oct 25 '18 at 16:31
R.C.CowsikR.C.Cowsik
31514
$endgroup$
add a comment |
$begingroup$
Yes. Take a basis ${e_i}$ of $V/W$, and choose $f_i$ in $V$ such that $f_i$ maps to $e_i$ under the quotient map. The map from $V/W$ to $V$ linearly extending the above gives an embedding. If $sum a_i e_i = z in V/W$ (a finite sum) then the image in $V$ would be $sum a_i f_i$. etc
edited Oct 25 '18 at 17:13
memerson
55510
answered Oct 25 '18 at 16:31
R.C.CowsikR.C.Cowsik
31514
$endgroup$
add a comment |
$begingroup$
Yes. Take a basis ${e_i}$ of $V/W$, and choose $f_i$ in $V$ such that $f_i$ maps to $e_i$ under the quotient map. The map from $V/W$ to $V$ linearly extending the above gives an embedding. If $sum a_i e_i = z in V/W$ (a finite sum) then the image in $V$ would be $sum a_i f_i$. etc
edited Oct 25 '18 at 17:13
memerson
55510
answered Oct 25 '18 at 16:31
R.C.CowsikR.C.Cowsik
31514
$endgroup$
Yes. Take a basis ${e_i}$ of $V/W$, and choose $f_i$ in $V$ such that $f_i$ maps to $e_i$ under the quotient map. The map from $V/W$ to $V$ linearly extending the above gives an embedding. If $sum a_i e_i = z in V/W$ (a finite sum) then the image in $V$ would be $sum a_i f_i$. etc
edited Oct 25 '18 at 17:13
memerson
55510
answered Oct 25 '18 at 16:31
R.C.CowsikR.C.Cowsik
31514
edited Oct 25 '18 at 17:13
memerson
55510
edited Oct 25 '18 at 17:13
memerson
55510
edited Oct 25 '18 at 17:13
memerson
55510
55510
answered Oct 25 '18 at 16:31
R.C.CowsikR.C.Cowsik
31514
answered Oct 25 '18 at 16:31
R.C.CowsikR.C.Cowsik
31514
answered Oct 25 '18 at 16:31
R.C.CowsikR.C.Cowsik
31514
31514
add a comment |
add a comment |
$begingroup$
I completely agree with the previous answers. However, please be aware that any such an embedding is actually a choice! So there is no canonical way on doing it (it is only canonical if you quotient out a trivial space, i.e. the whole space or 0). Furthermore, if you are interested, this property is usually referred to as semi-simplicity of a category (something quite powerful).
and to get back to Rhythmink's answer: an example for rings or groups were this is not possible is the map $$mathbb{Z} to mathbb{Z}/(p).$$
I leave it to you to show that this admits no leftinverse ,i.e. an embedding
$$mathbb{Z}/(p) to mathbb{Z}.$$
answered Oct 26 '18 at 8:42
EnkiduEnkidu
1,32619
$endgroup$
add a comment |
$begingroup$
I completely agree with the previous answers. However, please be aware that any such an embedding is actually a choice! So there is no canonical way on doing it (it is only canonical if you quotient out a trivial space, i.e. the whole space or 0). Furthermore, if you are interested, this property is usually referred to as semi-simplicity of a category (something quite powerful).
and to get back to Rhythmink's answer: an example for rings or groups were this is not possible is the map $$mathbb{Z} to mathbb{Z}/(p).$$
I leave it to you to show that this admits no leftinverse ,i.e. an embedding
$$mathbb{Z}/(p) to mathbb{Z}.$$
answered Oct 26 '18 at 8:42
EnkiduEnkidu
1,32619
$endgroup$
add a comment |
$begingroup$
I completely agree with the previous answers. However, please be aware that any such an embedding is actually a choice! So there is no canonical way on doing it (it is only canonical if you quotient out a trivial space, i.e. the whole space or 0). Furthermore, if you are interested, this property is usually referred to as semi-simplicity of a category (something quite powerful).
and to get back to Rhythmink's answer: an example for rings or groups were this is not possible is the map $$mathbb{Z} to mathbb{Z}/(p).$$
I leave it to you to show that this admits no leftinverse ,i.e. an embedding
$$mathbb{Z}/(p) to mathbb{Z}.$$
answered Oct 26 '18 at 8:42
EnkiduEnkidu
1,32619
$endgroup$
I completely agree with the previous answers. However, please be aware that any such an embedding is actually a choice! So there is no canonical way on doing it (it is only canonical if you quotient out a trivial space, i.e. the whole space or 0). Furthermore, if you are interested, this property is usually referred to as semi-simplicity of a category (something quite powerful).
and to get back to Rhythmink's answer: an example for rings or groups were this is not possible is the map $$mathbb{Z} to mathbb{Z}/(p).$$
I leave it to you to show that this admits no leftinverse ,i.e. an embedding
$$mathbb{Z}/(p) to mathbb{Z}.$$
answered Oct 26 '18 at 8:42
EnkiduEnkidu
1,32619
edited Dec 6 '18 at 8:09
answered Oct 26 '18 at 8:42
EnkiduEnkidu
1,32619
answered Oct 26 '18 at 8:42
EnkiduEnkidu
1,32619
answered Oct 26 '18 at 8:42
EnkiduEnkidu
1,32619
1,32619
add a comment |
add a comment |
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$begingroup$
The last assertion is false: For general finite-dimensional real or complex vector spaces there is no notion of orthogonality---this requires additional data (e.g., an inner product or Hermitian product).
$endgroup$
– Travis
Oct 26 '18 at 13:00