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What is the “right†extra information needed to define the trace of a map between two different vector spaces?
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Let $V,W$ be real vector spaces of dimension $d$, and let $T in textHom(V,W)$.
Is there a "natural piece of additional information" required in order to give a meaningful interpretation of the trace of $T$? (which is "less than" choosing an isomorphism $V cong W$).
Let me explain a bit what do I mean by "additional information" through other examples:
Suppose we want to define the determinant of $T$. Then sufficient additional information is a choice of preferred volume forms on $V,W$. Assuming we are given such forms, we can then define $det T$ by requiring
$$T^*(textVol_W)=det T cdot textVol_V.$$
Perhaps an even more economical option would be to provide an isomorphism $bigwedge^d V simeq bigwedge^d W$. Since $bigwedge^d T:bigwedge^d V to bigwedge^d W
$, by composing it with the given isomorphism we can identify $bigwedge^d T$ with a map $bigwedge^d V to bigwedge^d V$ which can then be naturally identified with a scalar.
(Choosing a volume form is equivalent to choosing an isomorphism $bigwedge^d V cong mathbbR$).
Of course, we could be given an isomorphism $V simeq W$ and consider $T$ as a map $V to V$, but my point is that this is much more than we really need.
In the same spirit, in order to define the singular values of a map, we need to choose inner products on both spaces. (Which is again less information then choosing isomorphisms of $V,W$ to $mathbbR^d$).
The question is whether there is an "analogous" piece of extra structure which naturally fits into the definition of the trace. Of course, when $V=W$ (or when we are given some isomorphism between $V$ and $W$) we can define the trace of $T$ as a map $V to V$.
(A coordinate-free version would be using the natural identification $V^* otimes V cong textHom(V,V)$, and the contraction map $V^* otimes V to F$. Alternatively we can choose a basis, and use the fact the trace of a matrix is invariant under conjugation).
I am aware that it's not entirely clear whether the trace can be given a purely "geometric meaning" (in the sense that it's not invariant under isometries for instance). This is different than the situation with the determinant, or the singular values.
Edit:
In a slight contrast to the last contrast, here is some evidence that the trace is at least "partially" a geometric creature: The space of trace-free matrices is the tangent space at the identity to $SL_d$ (the group of volume preserving matrices). Thus, I am guessing that we shall need at least something like an isomorphism $bigwedge^d V simeq bigwedge^d W$ (in order to define a notion of volume preservation for maps). Using this isomorphism, we can now identify $bigwedge^d T$ as a map $bigwedge^d V to bigwedge^d V$, and define
$$
"SL"= , bigwedge^d T=textId_bigwedge^d V .
$$
The problem is that I am not sure if we can "single out" which element is the analog of the "identity map" among all these elements of our $"SL"$. Thus, I am not sure that a choice of an isomorphism $bigwedge^d V simeq bigwedge^d W$ would suffice to even define what is "trace=$0$"...
linear-algebra abstract-algebra soft-question category-theory trace
asked Aug 24 at 13:13
Asaf Shachar
4,7163835
add a comment |Â
up vote
11
down vote
favorite
Let $V,W$ be real vector spaces of dimension $d$, and let $T in textHom(V,W)$.
Is there a "natural piece of additional information" required in order to give a meaningful interpretation of the trace of $T$? (which is "less than" choosing an isomorphism $V cong W$).
Let me explain a bit what do I mean by "additional information" through other examples:
Suppose we want to define the determinant of $T$. Then sufficient additional information is a choice of preferred volume forms on $V,W$. Assuming we are given such forms, we can then define $det T$ by requiring
$$T^*(textVol_W)=det T cdot textVol_V.$$
Perhaps an even more economical option would be to provide an isomorphism $bigwedge^d V simeq bigwedge^d W$. Since $bigwedge^d T:bigwedge^d V to bigwedge^d W
$, by composing it with the given isomorphism we can identify $bigwedge^d T$ with a map $bigwedge^d V to bigwedge^d V$ which can then be naturally identified with a scalar.
(Choosing a volume form is equivalent to choosing an isomorphism $bigwedge^d V cong mathbbR$).
Of course, we could be given an isomorphism $V simeq W$ and consider $T$ as a map $V to V$, but my point is that this is much more than we really need.
In the same spirit, in order to define the singular values of a map, we need to choose inner products on both spaces. (Which is again less information then choosing isomorphisms of $V,W$ to $mathbbR^d$).
The question is whether there is an "analogous" piece of extra structure which naturally fits into the definition of the trace. Of course, when $V=W$ (or when we are given some isomorphism between $V$ and $W$) we can define the trace of $T$ as a map $V to V$.
(A coordinate-free version would be using the natural identification $V^* otimes V cong textHom(V,V)$, and the contraction map $V^* otimes V to F$. Alternatively we can choose a basis, and use the fact the trace of a matrix is invariant under conjugation).
I am aware that it's not entirely clear whether the trace can be given a purely "geometric meaning" (in the sense that it's not invariant under isometries for instance). This is different than the situation with the determinant, or the singular values.
Edit:
In a slight contrast to the last contrast, here is some evidence that the trace is at least "partially" a geometric creature: The space of trace-free matrices is the tangent space at the identity to $SL_d$ (the group of volume preserving matrices). Thus, I am guessing that we shall need at least something like an isomorphism $bigwedge^d V simeq bigwedge^d W$ (in order to define a notion of volume preservation for maps). Using this isomorphism, we can now identify $bigwedge^d T$ as a map $bigwedge^d V to bigwedge^d V$, and define
$$
"SL"= , bigwedge^d T=textId_bigwedge^d V .
$$
The problem is that I am not sure if we can "single out" which element is the analog of the "identity map" among all these elements of our $"SL"$. Thus, I am not sure that a choice of an isomorphism $bigwedge^d V simeq bigwedge^d W$ would suffice to even define what is "trace=$0$"...
linear-algebra abstract-algebra soft-question category-theory trace
asked Aug 24 at 13:13
Asaf Shachar
4,7163835
add a comment |Â
up vote
11
down vote
favorite
up vote
11
down vote
favorite
Let $V,W$ be real vector spaces of dimension $d$, and let $T in textHom(V,W)$.
Is there a "natural piece of additional information" required in order to give a meaningful interpretation of the trace of $T$? (which is "less than" choosing an isomorphism $V cong W$).
Let me explain a bit what do I mean by "additional information" through other examples:
Suppose we want to define the determinant of $T$. Then sufficient additional information is a choice of preferred volume forms on $V,W$. Assuming we are given such forms, we can then define $det T$ by requiring
$$T^*(textVol_W)=det T cdot textVol_V.$$
Perhaps an even more economical option would be to provide an isomorphism $bigwedge^d V simeq bigwedge^d W$. Since $bigwedge^d T:bigwedge^d V to bigwedge^d W
$, by composing it with the given isomorphism we can identify $bigwedge^d T$ with a map $bigwedge^d V to bigwedge^d V$ which can then be naturally identified with a scalar.
(Choosing a volume form is equivalent to choosing an isomorphism $bigwedge^d V cong mathbbR$).
Of course, we could be given an isomorphism $V simeq W$ and consider $T$ as a map $V to V$, but my point is that this is much more than we really need.
In the same spirit, in order to define the singular values of a map, we need to choose inner products on both spaces. (Which is again less information then choosing isomorphisms of $V,W$ to $mathbbR^d$).
The question is whether there is an "analogous" piece of extra structure which naturally fits into the definition of the trace. Of course, when $V=W$ (or when we are given some isomorphism between $V$ and $W$) we can define the trace of $T$ as a map $V to V$.
(A coordinate-free version would be using the natural identification $V^* otimes V cong textHom(V,V)$, and the contraction map $V^* otimes V to F$. Alternatively we can choose a basis, and use the fact the trace of a matrix is invariant under conjugation).
I am aware that it's not entirely clear whether the trace can be given a purely "geometric meaning" (in the sense that it's not invariant under isometries for instance). This is different than the situation with the determinant, or the singular values.
Edit:
In a slight contrast to the last contrast, here is some evidence that the trace is at least "partially" a geometric creature: The space of trace-free matrices is the tangent space at the identity to $SL_d$ (the group of volume preserving matrices). Thus, I am guessing that we shall need at least something like an isomorphism $bigwedge^d V simeq bigwedge^d W$ (in order to define a notion of volume preservation for maps). Using this isomorphism, we can now identify $bigwedge^d T$ as a map $bigwedge^d V to bigwedge^d V$, and define
$$
"SL"= , bigwedge^d T=textId_bigwedge^d V .
$$
The problem is that I am not sure if we can "single out" which element is the analog of the "identity map" among all these elements of our $"SL"$. Thus, I am not sure that a choice of an isomorphism $bigwedge^d V simeq bigwedge^d W$ would suffice to even define what is "trace=$0$"...
linear-algebra abstract-algebra soft-question category-theory trace
asked Aug 24 at 13:13
Asaf Shachar
4,7163835
Let $V,W$ be real vector spaces of dimension $d$, and let $T in textHom(V,W)$.
Is there a "natural piece of additional information" required in order to give a meaningful interpretation of the trace of $T$? (which is "less than" choosing an isomorphism $V cong W$).
Let me explain a bit what do I mean by "additional information" through other examples:
Suppose we want to define the determinant of $T$. Then sufficient additional information is a choice of preferred volume forms on $V,W$. Assuming we are given such forms, we can then define $det T$ by requiring
$$T^*(textVol_W)=det T cdot textVol_V.$$
Perhaps an even more economical option would be to provide an isomorphism $bigwedge^d V simeq bigwedge^d W$. Since $bigwedge^d T:bigwedge^d V to bigwedge^d W
$, by composing it with the given isomorphism we can identify $bigwedge^d T$ with a map $bigwedge^d V to bigwedge^d V$ which can then be naturally identified with a scalar.
(Choosing a volume form is equivalent to choosing an isomorphism $bigwedge^d V cong mathbbR$).
Of course, we could be given an isomorphism $V simeq W$ and consider $T$ as a map $V to V$, but my point is that this is much more than we really need.
In the same spirit, in order to define the singular values of a map, we need to choose inner products on both spaces. (Which is again less information then choosing isomorphisms of $V,W$ to $mathbbR^d$).
The question is whether there is an "analogous" piece of extra structure which naturally fits into the definition of the trace. Of course, when $V=W$ (or when we are given some isomorphism between $V$ and $W$) we can define the trace of $T$ as a map $V to V$.
(A coordinate-free version would be using the natural identification $V^* otimes V cong textHom(V,V)$, and the contraction map $V^* otimes V to F$. Alternatively we can choose a basis, and use the fact the trace of a matrix is invariant under conjugation).
I am aware that it's not entirely clear whether the trace can be given a purely "geometric meaning" (in the sense that it's not invariant under isometries for instance). This is different than the situation with the determinant, or the singular values.
Edit:
In a slight contrast to the last contrast, here is some evidence that the trace is at least "partially" a geometric creature: The space of trace-free matrices is the tangent space at the identity to $SL_d$ (the group of volume preserving matrices). Thus, I am guessing that we shall need at least something like an isomorphism $bigwedge^d V simeq bigwedge^d W$ (in order to define a notion of volume preservation for maps). Using this isomorphism, we can now identify $bigwedge^d T$ as a map $bigwedge^d V to bigwedge^d V$, and define
$$
"SL"= , bigwedge^d T=textId_bigwedge^d V .
$$
The problem is that I am not sure if we can "single out" which element is the analog of the "identity map" among all these elements of our $"SL"$. Thus, I am not sure that a choice of an isomorphism $bigwedge^d V simeq bigwedge^d W$ would suffice to even define what is "trace=$0$"...
linear-algebra abstract-algebra soft-question category-theory trace
asked Aug 24 at 13:13
Asaf Shachar
4,7163835
edited Aug 24 at 15:27
asked Aug 24 at 13:13
Asaf Shachar
4,7163835
asked Aug 24 at 13:13
Asaf Shachar
4,7163835
asked Aug 24 at 13:13
Asaf Shachar
4,7163835
4,7163835
add a comment |Â
add a comment |Â
1 Answer
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As you observe, if we choose an isomorphism $A:Wto V$, we can define the trace of $f:Vto W$ to be
$$mathrmTr_A(f):=mathrmTr(fA)=mathrmTr(A f).$$
If $B:Wto V$ is another isomorphism, then $B=MA$ for some $Min GL(V)$, and
$$
mathrmTr_B(f)=mathrmTr(MA f).$$
If $mathrmTr_A=mathrmTr_B$, then $mathrmTr(MC)=mathrmTr(C)$ for every $CinmathrmEnd(V)$. This implies $mathrmTr((M-mathrmId_V)C)=0$ for all $C$, which implies $M=mathrmId_V$. So the trace function determines the isomorphism from $W$ to $V$.
By contrast, $det(MA)=det(A)$ for all $A$ if and only if $det(M)=1$. So knowing the determinant function only defines an isomorphism of $W$ with $V$ modulo determinant $1$ maps. The orbit space $mathrmIso(W,V)/SL(V)$ can be identified with $mathrmIso(bigwedge^d W,bigwedge^d V)$.
answered Aug 24 at 16:12
Julian Rosen
11.4k12247
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
As you observe, if we choose an isomorphism $A:Wto V$, we can define the trace of $f:Vto W$ to be
$$mathrmTr_A(f):=mathrmTr(fA)=mathrmTr(A f).$$
If $B:Wto V$ is another isomorphism, then $B=MA$ for some $Min GL(V)$, and
$$
mathrmTr_B(f)=mathrmTr(MA f).$$
If $mathrmTr_A=mathrmTr_B$, then $mathrmTr(MC)=mathrmTr(C)$ for every $CinmathrmEnd(V)$. This implies $mathrmTr((M-mathrmId_V)C)=0$ for all $C$, which implies $M=mathrmId_V$. So the trace function determines the isomorphism from $W$ to $V$.
By contrast, $det(MA)=det(A)$ for all $A$ if and only if $det(M)=1$. So knowing the determinant function only defines an isomorphism of $W$ with $V$ modulo determinant $1$ maps. The orbit space $mathrmIso(W,V)/SL(V)$ can be identified with $mathrmIso(bigwedge^d W,bigwedge^d V)$.
answered Aug 24 at 16:12
Julian Rosen
11.4k12247
add a comment |Â
up vote
5
down vote
accepted
As you observe, if we choose an isomorphism $A:Wto V$, we can define the trace of $f:Vto W$ to be
$$mathrmTr_A(f):=mathrmTr(fA)=mathrmTr(A f).$$
If $B:Wto V$ is another isomorphism, then $B=MA$ for some $Min GL(V)$, and
$$
mathrmTr_B(f)=mathrmTr(MA f).$$
If $mathrmTr_A=mathrmTr_B$, then $mathrmTr(MC)=mathrmTr(C)$ for every $CinmathrmEnd(V)$. This implies $mathrmTr((M-mathrmId_V)C)=0$ for all $C$, which implies $M=mathrmId_V$. So the trace function determines the isomorphism from $W$ to $V$.
By contrast, $det(MA)=det(A)$ for all $A$ if and only if $det(M)=1$. So knowing the determinant function only defines an isomorphism of $W$ with $V$ modulo determinant $1$ maps. The orbit space $mathrmIso(W,V)/SL(V)$ can be identified with $mathrmIso(bigwedge^d W,bigwedge^d V)$.
answered Aug 24 at 16:12
Julian Rosen
11.4k12247
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
As you observe, if we choose an isomorphism $A:Wto V$, we can define the trace of $f:Vto W$ to be
$$mathrmTr_A(f):=mathrmTr(fA)=mathrmTr(A f).$$
If $B:Wto V$ is another isomorphism, then $B=MA$ for some $Min GL(V)$, and
$$
mathrmTr_B(f)=mathrmTr(MA f).$$
If $mathrmTr_A=mathrmTr_B$, then $mathrmTr(MC)=mathrmTr(C)$ for every $CinmathrmEnd(V)$. This implies $mathrmTr((M-mathrmId_V)C)=0$ for all $C$, which implies $M=mathrmId_V$. So the trace function determines the isomorphism from $W$ to $V$.
By contrast, $det(MA)=det(A)$ for all $A$ if and only if $det(M)=1$. So knowing the determinant function only defines an isomorphism of $W$ with $V$ modulo determinant $1$ maps. The orbit space $mathrmIso(W,V)/SL(V)$ can be identified with $mathrmIso(bigwedge^d W,bigwedge^d V)$.
answered Aug 24 at 16:12
Julian Rosen
11.4k12247
As you observe, if we choose an isomorphism $A:Wto V$, we can define the trace of $f:Vto W$ to be
$$mathrmTr_A(f):=mathrmTr(fA)=mathrmTr(A f).$$
If $B:Wto V$ is another isomorphism, then $B=MA$ for some $Min GL(V)$, and
$$
mathrmTr_B(f)=mathrmTr(MA f).$$
If $mathrmTr_A=mathrmTr_B$, then $mathrmTr(MC)=mathrmTr(C)$ for every $CinmathrmEnd(V)$. This implies $mathrmTr((M-mathrmId_V)C)=0$ for all $C$, which implies $M=mathrmId_V$. So the trace function determines the isomorphism from $W$ to $V$.
By contrast, $det(MA)=det(A)$ for all $A$ if and only if $det(M)=1$. So knowing the determinant function only defines an isomorphism of $W$ with $V$ modulo determinant $1$ maps. The orbit space $mathrmIso(W,V)/SL(V)$ can be identified with $mathrmIso(bigwedge^d W,bigwedge^d V)$.
answered Aug 24 at 16:12
Julian Rosen
11.4k12247
answered Aug 24 at 16:12
Julian Rosen
11.4k12247
answered Aug 24 at 16:12
Julian Rosen
11.4k12247
answered Aug 24 at 16:12
Julian Rosen
11.4k12247
11.4k12247
add a comment |Â
add a comment |Â
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