Non-standard tensor products of inner product spacesLinearity of the inner product using the parallelogram lawInner product of linear bounded operators between Hilbert spacesA doubt about tensor product on Hilbert SpacesInner product spaces without symmetry/hermitian axiomExistence of $SO(n)$-isotropic inner products which are not $O(n)$-isotropicDifferent inner products for vector spaces of random variablesCharacterizing (minimal) tensor product inside Hilbert C*-moduleInner Product on tensor product of Hilbert spaces is unique?The tensor product of two bounded operatorsWhich inner products preserve positive correlation?
Non-standard tensor products of inner product spaces
Linearity of the inner product using the parallelogram lawInner product of linear bounded operators between Hilbert spacesA doubt about tensor product on Hilbert SpacesInner product spaces without symmetry/hermitian axiomExistence of $SO(n)$-isotropic inner products which are not $O(n)$-isotropicDifferent inner products for vector spaces of random variablesCharacterizing (minimal) tensor product inside Hilbert C*-moduleInner Product on tensor product of Hilbert spaces is unique?The tensor product of two bounded operatorsWhich inner products preserve positive correlation?
$begingroup$
For two inner product spaces $(mathcalV, (cdot,cdot)_V)$ and $(mathcalW, (cdot,cdot)_W)$, we can put an inner product on their tensor product in the obvious way:
$$
(1) ~~~~ langle v otimes w, v' otimes w'rangle := langle v,v'rangle_V langle w,w'rangle_W.
$$
This then implies that
$$
(2) ~~~~ |v otimes w| = |v|_V |w|_W.
$$
Is there an example of an inner product on $mathcalV otimes mathcalW$ such that (2) holds, but the inner product is not of the form (1)? Are such things of interest? Do they have a name?
fa.functional-analysis oa.operator-algebras hilbert-spaces inner-product
asked May 30 at 22:24
Pierre DuboisPierre Dubois
1888
$endgroup$
This question has an open bounty worth +200
reputation from Pierre Dubois ending ending at 2019-06-12 00:59:26Z">in 6 days.
Looking for an answer drawing from credible and/or official sources.
|
show 7 more comments
$begingroup$
For two inner product spaces $(mathcalV, (cdot,cdot)_V)$ and $(mathcalW, (cdot,cdot)_W)$, we can put an inner product on their tensor product in the obvious way:
$$
(1) ~~~~ langle v otimes w, v' otimes w'rangle := langle v,v'rangle_V langle w,w'rangle_W.
$$
This then implies that
$$
(2) ~~~~ |v otimes w| = |v|_V |w|_W.
$$
Is there an example of an inner product on $mathcalV otimes mathcalW$ such that (2) holds, but the inner product is not of the form (1)? Are such things of interest? Do they have a name?
fa.functional-analysis oa.operator-algebras hilbert-spaces inner-product
asked May 30 at 22:24
Pierre DuboisPierre Dubois
1888
$endgroup$
This question has an open bounty worth +200
reputation from Pierre Dubois ending ending at 2019-06-12 00:59:26Z">in 6 days.
Looking for an answer drawing from credible and/or official sources.
1
$begingroup$
However, there are various different norms on the tensor products of Hilbert spaces which do not come from an inner product (i.e. they do not satisfy the parallelogram law).
$endgroup$
– Matthias Ludewig
May 31 at 3:59
$begingroup$
There is a construction called the diamond norm that I think goes close to it but maybe not entirely what you want.
$endgroup$
– lcv
May 31 at 4:22
$begingroup$
@lcv It doesn't come from an inner product, and the OP is asking about inner products
$endgroup$
– Yemon Choi
May 31 at 12:03
1
$begingroup$
To be clear - as Yemon Choi says - I am assuming that the norm in (2) is induced by an inner product.
$endgroup$
– Pierre Dubois
May 31 at 12:39
$begingroup$
@PierreDubois no problem yes now it's clearer I can remove my answer.
$endgroup$
– lcv
May 31 at 15:14
|
show 7 more comments
$begingroup$
For two inner product spaces $(mathcalV, (cdot,cdot)_V)$ and $(mathcalW, (cdot,cdot)_W)$, we can put an inner product on their tensor product in the obvious way:
$$
(1) ~~~~ langle v otimes w, v' otimes w'rangle := langle v,v'rangle_V langle w,w'rangle_W.
$$
This then implies that
$$
(2) ~~~~ |v otimes w| = |v|_V |w|_W.
$$
Is there an example of an inner product on $mathcalV otimes mathcalW$ such that (2) holds, but the inner product is not of the form (1)? Are such things of interest? Do they have a name?
fa.functional-analysis oa.operator-algebras hilbert-spaces inner-product
asked May 30 at 22:24
Pierre DuboisPierre Dubois
1888
$endgroup$
For two inner product spaces $(mathcalV, (cdot,cdot)_V)$ and $(mathcalW, (cdot,cdot)_W)$, we can put an inner product on their tensor product in the obvious way:
$$
(1) ~~~~ langle v otimes w, v' otimes w'rangle := langle v,v'rangle_V langle w,w'rangle_W.
$$
This then implies that
$$
(2) ~~~~ |v otimes w| = |v|_V |w|_W.
$$
Is there an example of an inner product on $mathcalV otimes mathcalW$ such that (2) holds, but the inner product is not of the form (1)? Are such things of interest? Do they have a name?
fa.functional-analysis oa.operator-algebras hilbert-spaces inner-product
fa.functional-analysis oa.operator-algebras hilbert-spaces inner-product
asked May 30 at 22:24
Pierre DuboisPierre Dubois
1888
asked May 30 at 22:24
Pierre DuboisPierre Dubois
1888
asked May 30 at 22:24
Pierre DuboisPierre Dubois
1888
asked May 30 at 22:24
Pierre DuboisPierre Dubois
1888
asked May 30 at 22:24
Pierre DuboisPierre Dubois
1888
1888
This question has an open bounty worth +200
reputation from Pierre Dubois ending ending at 2019-06-12 00:59:26Z">in 6 days.
Looking for an answer drawing from credible and/or official sources.
This question has an open bounty worth +200
reputation from Pierre Dubois ending ending at 2019-06-12 00:59:26Z">in 6 days.
Looking for an answer drawing from credible and/or official sources.
1
$begingroup$
However, there are various different norms on the tensor products of Hilbert spaces which do not come from an inner product (i.e. they do not satisfy the parallelogram law).
$endgroup$
– Matthias Ludewig
May 31 at 3:59
$begingroup$
There is a construction called the diamond norm that I think goes close to it but maybe not entirely what you want.
$endgroup$
– lcv
May 31 at 4:22
$begingroup$
@lcv It doesn't come from an inner product, and the OP is asking about inner products
$endgroup$
– Yemon Choi
May 31 at 12:03
1
$begingroup$
To be clear - as Yemon Choi says - I am assuming that the norm in (2) is induced by an inner product.
$endgroup$
– Pierre Dubois
May 31 at 12:39
$begingroup$
@PierreDubois no problem yes now it's clearer I can remove my answer.
$endgroup$
– lcv
May 31 at 15:14
|
show 7 more comments
1
$begingroup$
However, there are various different norms on the tensor products of Hilbert spaces which do not come from an inner product (i.e. they do not satisfy the parallelogram law).
$endgroup$
– Matthias Ludewig
May 31 at 3:59
$begingroup$
There is a construction called the diamond norm that I think goes close to it but maybe not entirely what you want.
$endgroup$
– lcv
May 31 at 4:22
$begingroup$
@lcv It doesn't come from an inner product, and the OP is asking about inner products
$endgroup$
– Yemon Choi
May 31 at 12:03
1
$begingroup$
To be clear - as Yemon Choi says - I am assuming that the norm in (2) is induced by an inner product.
$endgroup$
– Pierre Dubois
May 31 at 12:39
$begingroup$
@PierreDubois no problem yes now it's clearer I can remove my answer.
$endgroup$
– lcv
May 31 at 15:14
1
1
$begingroup$
However, there are various different norms on the tensor products of Hilbert spaces which do not come from an inner product (i.e. they do not satisfy the parallelogram law).
$endgroup$
– Matthias Ludewig
May 31 at 3:59
$begingroup$
However, there are various different norms on the tensor products of Hilbert spaces which do not come from an inner product (i.e. they do not satisfy the parallelogram law).
$endgroup$
– Matthias Ludewig
May 31 at 3:59
$begingroup$
There is a construction called the diamond norm that I think goes close to it but maybe not entirely what you want.
$endgroup$
– lcv
May 31 at 4:22
$begingroup$
There is a construction called the diamond norm that I think goes close to it but maybe not entirely what you want.
$endgroup$
– lcv
May 31 at 4:22
$begingroup$
@lcv It doesn't come from an inner product, and the OP is asking about inner products
$endgroup$
– Yemon Choi
May 31 at 12:03
$begingroup$
@lcv It doesn't come from an inner product, and the OP is asking about inner products
$endgroup$
– Yemon Choi
May 31 at 12:03
1
1
$begingroup$
To be clear - as Yemon Choi says - I am assuming that the norm in (2) is induced by an inner product.
$endgroup$
– Pierre Dubois
May 31 at 12:39
$begingroup$
To be clear - as Yemon Choi says - I am assuming that the norm in (2) is induced by an inner product.
$endgroup$
– Pierre Dubois
May 31 at 12:39
$begingroup$
@PierreDubois no problem yes now it's clearer I can remove my answer.
$endgroup$
– lcv
May 31 at 15:14
$begingroup$
@PierreDubois no problem yes now it's clearer I can remove my answer.
$endgroup$
– lcv
May 31 at 15:14
|
show 7 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Assuming complex scalars, no, any inner product which satisfies (2) for all $v in mathcalV$ and $w in mathcalW$ has the given form. To see this, let $[cdot,cdot]$ be any inner product which satisfies (2). So right away we know that $[votimes w,votimes w] = langle votimes w, votimes wrangle$ for all $v$ and $w$. Next, for $v,v' in mathcalV$ and $w in mathcalW$, we know that $$[(v + v')otimes w, (v + v')otimes w] = langle (v + v')otimes w, (v + v')otimes wrangle.$$ Expanding this out and applying $[votimes w,votimes w] = langle votimes w,votimes wrangle$ plus the same for $v'otimes w$ yields $2rm Re[votimes w, v'otimes w] = 2rm Relangle votimes w, v'otimes wrangle$. Replacing $v$ with $iv$ yields equality of the imaginary parts too. So now we know that $[votimes w,v'otimes w] = langle votimes w,v'otimes wrangle$ for all $v$, $v'$, and $w$.
Finally, for any $v,v' in mathcalV$ and $w,w' in mathcalW$ we have $$[(v + v')otimes (w + w'), (v + v')otimes (w + w')] = langle (v + v')otimes (w + w'), (v + v')otimes (w + w')rangle.$$ Expanding this out and cancelling all the equalities we already have then leaves only $$2rm Re([votimes w, v'otimes w'] + [votimes w', v'otimes w]) = 2rm Re(langle votimes w, v'otimes w'rangle + langle votimes w', v'otimes wrangle).$$ Replacing $v$ with $iv$ yields equality of the imaginary parts too, so that $$[votimes w, v'otimes w'] + [votimes w', v'otimes w] = langle votimes w, v'otimes w'rangle + langle votimes w', v'otimes wrangle.$$ Then replacing $w'$ with $iw'$ yields $$[votimes w, v'otimes w'] - [votimes w', v'otimes w] = langle votimes w, v'otimes w'rangle - langle votimes w', v'otimes wrangle,$$ so that $[votimes w, v'otimes w'] = langle votimes w,v'otimes w'rangle$. As every element of the algebraic tensor product $mathcalVotimesmathcalW$ is a linear combination of elementary tensors, this shows that $[cdot,cdot] = langlecdot,cdotrangle$.
I feel there ought to be a one-line proof of this, but I don't quite see it.
answered May 31 at 3:29
Nik WeaverNik Weaver
23k152136
$endgroup$
add a comment |
$begingroup$
Let $V, W$ be 2-dimensional with orthonormal bases $v_0, v_1$ and $w_0, w_1$. If you expand out your condition (2), you obtain the following conditions:
$$ langle v_i otimes w_j, v_i otimes w_j rangle = 1 \
langle v_i otimes w_j, v_i otimes w_k rangle = langle v_i otimes w_j, v_k otimes w_j rangle = 0 \
langle v_0 otimes w_0, v_1 otimes w_1 rangle + langle v_1 otimes w_0, v_0 otimes w_1 rangle = 0
$$
Clearly you can pick any value for $langle v_0 otimes w_0, v_1 otimes w_1 rangle$. Any nonzero choice will not satisfy your condition (1), since the RHS will be 0. So it is certainly possible. I'm not sure if they're of interest.
EDIT: Since there's disagreement in the comments, I'm posting my calculation. Consider general vectors $a_0 v_0 + a_1 v_1 in V$, $b_0 w_0 + b_1 w_1 in W$. Let's mangle Einstein notation by putting $langle v_i otimes w_j, v_k otimes w_lrangle = g_ij^kl$.
(2) says that $ langle a_0 v_0 + a_1 v_1, a_0 v_0 + a_1 v_1 rangle = (a_0^2 + a_1^2)(b_0^2 + b_1^2)$. Expanding out the LHS gives
$$ a_0^2 b_0^2 g_00^00 + a_1^2 b_0^2 g_10^10 + a_0^2 b_1^2 g_01^01 + a_1^2 b_1^2 g_11^11 + \
2left(a_1 a_0 b_0^2 g_10^00 + a_0^2 b_1 b_0 g_01^00 + a_1^2 b_1 b_0 g^10_11 + a_1 a_0 b_1^2 g^01_11right) + \
4a_1 a_0 b_1 b_0 left(g^00_11 + g^10_01 right).$$
By choosing unit vectors, it is clear that $g^ij_ij = 1$. So the first line cancels with the RHS. By choosing a sum of two unit vectors, it is clear that $g^ij_ik = g^ij_kj = 0$, so the second line vanishes. We are left with $4a_1 a_0 b_1 b_0 (g^00_11 + g^10_01)$ = 0. If we choose our $g$ so that $g^00_11 + g^10_01 = 0$, this quantity will always vanish, and so (2) will hold for any pair of vectors.
answered May 31 at 0:52
Kevin CastoKevin Casto
1,21421213
$endgroup$
$begingroup$
This isn't correct; you've only assumed (2) for basis vectors, not all $v in mathcalV$ and $w in mathcalW$.
$endgroup$
– Nik Weaver
May 31 at 3:06
1
$begingroup$
Hi @NikWeaver -- I've posted my calculation, let me know if there's a mistake. Thanks!
$endgroup$
– Kevin Casto
May 31 at 3:31
1
$begingroup$
At first glance it looks right (assuming real scalars), but I also think my answer is right ... going to have to think about this for a minute.
$endgroup$
– Nik Weaver
May 31 at 3:32
2
$begingroup$
@NikWeaver Heuristic answer: $Sym^2(V^* otimes W^*) = Sym^2(V^*) otimes Sym^2(W^*) + Lambda^2(V^*) otimes Lambda^2(W^*)$. This is why, when you fix the inner product on $V$ and $W$, you get something antisymmetric in $V$ and in $W$ (as in my answer). I'm trying to make this precise in terms of kernels vs projections, but in any case counting constants should tell you that (2) doesn't pin down what the total inner product is.
$endgroup$
– Kevin Casto
May 31 at 3:40
2
$begingroup$
Actually, the scalars might be the issue --- my argument for uniqueness uses complex numbers. So we get different answers depending on whether we use real or complex scalars?
$endgroup$
– Nik Weaver
May 31 at 3:41
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Assuming complex scalars, no, any inner product which satisfies (2) for all $v in mathcalV$ and $w in mathcalW$ has the given form. To see this, let $[cdot,cdot]$ be any inner product which satisfies (2). So right away we know that $[votimes w,votimes w] = langle votimes w, votimes wrangle$ for all $v$ and $w$. Next, for $v,v' in mathcalV$ and $w in mathcalW$, we know that $$[(v + v')otimes w, (v + v')otimes w] = langle (v + v')otimes w, (v + v')otimes wrangle.$$ Expanding this out and applying $[votimes w,votimes w] = langle votimes w,votimes wrangle$ plus the same for $v'otimes w$ yields $2rm Re[votimes w, v'otimes w] = 2rm Relangle votimes w, v'otimes wrangle$. Replacing $v$ with $iv$ yields equality of the imaginary parts too. So now we know that $[votimes w,v'otimes w] = langle votimes w,v'otimes wrangle$ for all $v$, $v'$, and $w$.
Finally, for any $v,v' in mathcalV$ and $w,w' in mathcalW$ we have $$[(v + v')otimes (w + w'), (v + v')otimes (w + w')] = langle (v + v')otimes (w + w'), (v + v')otimes (w + w')rangle.$$ Expanding this out and cancelling all the equalities we already have then leaves only $$2rm Re([votimes w, v'otimes w'] + [votimes w', v'otimes w]) = 2rm Re(langle votimes w, v'otimes w'rangle + langle votimes w', v'otimes wrangle).$$ Replacing $v$ with $iv$ yields equality of the imaginary parts too, so that $$[votimes w, v'otimes w'] + [votimes w', v'otimes w] = langle votimes w, v'otimes w'rangle + langle votimes w', v'otimes wrangle.$$ Then replacing $w'$ with $iw'$ yields $$[votimes w, v'otimes w'] - [votimes w', v'otimes w] = langle votimes w, v'otimes w'rangle - langle votimes w', v'otimes wrangle,$$ so that $[votimes w, v'otimes w'] = langle votimes w,v'otimes w'rangle$. As every element of the algebraic tensor product $mathcalVotimesmathcalW$ is a linear combination of elementary tensors, this shows that $[cdot,cdot] = langlecdot,cdotrangle$.
I feel there ought to be a one-line proof of this, but I don't quite see it.
answered May 31 at 3:29
Nik WeaverNik Weaver
23k152136
$endgroup$
add a comment |
$begingroup$
Assuming complex scalars, no, any inner product which satisfies (2) for all $v in mathcalV$ and $w in mathcalW$ has the given form. To see this, let $[cdot,cdot]$ be any inner product which satisfies (2). So right away we know that $[votimes w,votimes w] = langle votimes w, votimes wrangle$ for all $v$ and $w$. Next, for $v,v' in mathcalV$ and $w in mathcalW$, we know that $$[(v + v')otimes w, (v + v')otimes w] = langle (v + v')otimes w, (v + v')otimes wrangle.$$ Expanding this out and applying $[votimes w,votimes w] = langle votimes w,votimes wrangle$ plus the same for $v'otimes w$ yields $2rm Re[votimes w, v'otimes w] = 2rm Relangle votimes w, v'otimes wrangle$. Replacing $v$ with $iv$ yields equality of the imaginary parts too. So now we know that $[votimes w,v'otimes w] = langle votimes w,v'otimes wrangle$ for all $v$, $v'$, and $w$.
Finally, for any $v,v' in mathcalV$ and $w,w' in mathcalW$ we have $$[(v + v')otimes (w + w'), (v + v')otimes (w + w')] = langle (v + v')otimes (w + w'), (v + v')otimes (w + w')rangle.$$ Expanding this out and cancelling all the equalities we already have then leaves only $$2rm Re([votimes w, v'otimes w'] + [votimes w', v'otimes w]) = 2rm Re(langle votimes w, v'otimes w'rangle + langle votimes w', v'otimes wrangle).$$ Replacing $v$ with $iv$ yields equality of the imaginary parts too, so that $$[votimes w, v'otimes w'] + [votimes w', v'otimes w] = langle votimes w, v'otimes w'rangle + langle votimes w', v'otimes wrangle.$$ Then replacing $w'$ with $iw'$ yields $$[votimes w, v'otimes w'] - [votimes w', v'otimes w] = langle votimes w, v'otimes w'rangle - langle votimes w', v'otimes wrangle,$$ so that $[votimes w, v'otimes w'] = langle votimes w,v'otimes w'rangle$. As every element of the algebraic tensor product $mathcalVotimesmathcalW$ is a linear combination of elementary tensors, this shows that $[cdot,cdot] = langlecdot,cdotrangle$.
I feel there ought to be a one-line proof of this, but I don't quite see it.
answered May 31 at 3:29
Nik WeaverNik Weaver
23k152136
$endgroup$
add a comment |
$begingroup$
Assuming complex scalars, no, any inner product which satisfies (2) for all $v in mathcalV$ and $w in mathcalW$ has the given form. To see this, let $[cdot,cdot]$ be any inner product which satisfies (2). So right away we know that $[votimes w,votimes w] = langle votimes w, votimes wrangle$ for all $v$ and $w$. Next, for $v,v' in mathcalV$ and $w in mathcalW$, we know that $$[(v + v')otimes w, (v + v')otimes w] = langle (v + v')otimes w, (v + v')otimes wrangle.$$ Expanding this out and applying $[votimes w,votimes w] = langle votimes w,votimes wrangle$ plus the same for $v'otimes w$ yields $2rm Re[votimes w, v'otimes w] = 2rm Relangle votimes w, v'otimes wrangle$. Replacing $v$ with $iv$ yields equality of the imaginary parts too. So now we know that $[votimes w,v'otimes w] = langle votimes w,v'otimes wrangle$ for all $v$, $v'$, and $w$.
Finally, for any $v,v' in mathcalV$ and $w,w' in mathcalW$ we have $$[(v + v')otimes (w + w'), (v + v')otimes (w + w')] = langle (v + v')otimes (w + w'), (v + v')otimes (w + w')rangle.$$ Expanding this out and cancelling all the equalities we already have then leaves only $$2rm Re([votimes w, v'otimes w'] + [votimes w', v'otimes w]) = 2rm Re(langle votimes w, v'otimes w'rangle + langle votimes w', v'otimes wrangle).$$ Replacing $v$ with $iv$ yields equality of the imaginary parts too, so that $$[votimes w, v'otimes w'] + [votimes w', v'otimes w] = langle votimes w, v'otimes w'rangle + langle votimes w', v'otimes wrangle.$$ Then replacing $w'$ with $iw'$ yields $$[votimes w, v'otimes w'] - [votimes w', v'otimes w] = langle votimes w, v'otimes w'rangle - langle votimes w', v'otimes wrangle,$$ so that $[votimes w, v'otimes w'] = langle votimes w,v'otimes w'rangle$. As every element of the algebraic tensor product $mathcalVotimesmathcalW$ is a linear combination of elementary tensors, this shows that $[cdot,cdot] = langlecdot,cdotrangle$.
I feel there ought to be a one-line proof of this, but I don't quite see it.
answered May 31 at 3:29
Nik WeaverNik Weaver
23k152136
$endgroup$
Assuming complex scalars, no, any inner product which satisfies (2) for all $v in mathcalV$ and $w in mathcalW$ has the given form. To see this, let $[cdot,cdot]$ be any inner product which satisfies (2). So right away we know that $[votimes w,votimes w] = langle votimes w, votimes wrangle$ for all $v$ and $w$. Next, for $v,v' in mathcalV$ and $w in mathcalW$, we know that $$[(v + v')otimes w, (v + v')otimes w] = langle (v + v')otimes w, (v + v')otimes wrangle.$$ Expanding this out and applying $[votimes w,votimes w] = langle votimes w,votimes wrangle$ plus the same for $v'otimes w$ yields $2rm Re[votimes w, v'otimes w] = 2rm Relangle votimes w, v'otimes wrangle$. Replacing $v$ with $iv$ yields equality of the imaginary parts too. So now we know that $[votimes w,v'otimes w] = langle votimes w,v'otimes wrangle$ for all $v$, $v'$, and $w$.
Finally, for any $v,v' in mathcalV$ and $w,w' in mathcalW$ we have $$[(v + v')otimes (w + w'), (v + v')otimes (w + w')] = langle (v + v')otimes (w + w'), (v + v')otimes (w + w')rangle.$$ Expanding this out and cancelling all the equalities we already have then leaves only $$2rm Re([votimes w, v'otimes w'] + [votimes w', v'otimes w]) = 2rm Re(langle votimes w, v'otimes w'rangle + langle votimes w', v'otimes wrangle).$$ Replacing $v$ with $iv$ yields equality of the imaginary parts too, so that $$[votimes w, v'otimes w'] + [votimes w', v'otimes w] = langle votimes w, v'otimes w'rangle + langle votimes w', v'otimes wrangle.$$ Then replacing $w'$ with $iw'$ yields $$[votimes w, v'otimes w'] - [votimes w', v'otimes w] = langle votimes w, v'otimes w'rangle - langle votimes w', v'otimes wrangle,$$ so that $[votimes w, v'otimes w'] = langle votimes w,v'otimes w'rangle$. As every element of the algebraic tensor product $mathcalVotimesmathcalW$ is a linear combination of elementary tensors, this shows that $[cdot,cdot] = langlecdot,cdotrangle$.
I feel there ought to be a one-line proof of this, but I don't quite see it.
answered May 31 at 3:29
Nik WeaverNik Weaver
23k152136
edited May 31 at 4:02
answered May 31 at 3:29
Nik WeaverNik Weaver
23k152136
answered May 31 at 3:29
Nik WeaverNik Weaver
23k152136
answered May 31 at 3:29
Nik WeaverNik Weaver
23k152136
23k152136
add a comment |
add a comment |
$begingroup$
Let $V, W$ be 2-dimensional with orthonormal bases $v_0, v_1$ and $w_0, w_1$. If you expand out your condition (2), you obtain the following conditions:
$$ langle v_i otimes w_j, v_i otimes w_j rangle = 1 \
langle v_i otimes w_j, v_i otimes w_k rangle = langle v_i otimes w_j, v_k otimes w_j rangle = 0 \
langle v_0 otimes w_0, v_1 otimes w_1 rangle + langle v_1 otimes w_0, v_0 otimes w_1 rangle = 0
$$
Clearly you can pick any value for $langle v_0 otimes w_0, v_1 otimes w_1 rangle$. Any nonzero choice will not satisfy your condition (1), since the RHS will be 0. So it is certainly possible. I'm not sure if they're of interest.
EDIT: Since there's disagreement in the comments, I'm posting my calculation. Consider general vectors $a_0 v_0 + a_1 v_1 in V$, $b_0 w_0 + b_1 w_1 in W$. Let's mangle Einstein notation by putting $langle v_i otimes w_j, v_k otimes w_lrangle = g_ij^kl$.
(2) says that $ langle a_0 v_0 + a_1 v_1, a_0 v_0 + a_1 v_1 rangle = (a_0^2 + a_1^2)(b_0^2 + b_1^2)$. Expanding out the LHS gives
$$ a_0^2 b_0^2 g_00^00 + a_1^2 b_0^2 g_10^10 + a_0^2 b_1^2 g_01^01 + a_1^2 b_1^2 g_11^11 + \
2left(a_1 a_0 b_0^2 g_10^00 + a_0^2 b_1 b_0 g_01^00 + a_1^2 b_1 b_0 g^10_11 + a_1 a_0 b_1^2 g^01_11right) + \
4a_1 a_0 b_1 b_0 left(g^00_11 + g^10_01 right).$$
By choosing unit vectors, it is clear that $g^ij_ij = 1$. So the first line cancels with the RHS. By choosing a sum of two unit vectors, it is clear that $g^ij_ik = g^ij_kj = 0$, so the second line vanishes. We are left with $4a_1 a_0 b_1 b_0 (g^00_11 + g^10_01)$ = 0. If we choose our $g$ so that $g^00_11 + g^10_01 = 0$, this quantity will always vanish, and so (2) will hold for any pair of vectors.
answered May 31 at 0:52
Kevin CastoKevin Casto
1,21421213
$endgroup$
$begingroup$
This isn't correct; you've only assumed (2) for basis vectors, not all $v in mathcalV$ and $w in mathcalW$.
$endgroup$
– Nik Weaver
May 31 at 3:06
1
$begingroup$
Hi @NikWeaver -- I've posted my calculation, let me know if there's a mistake. Thanks!
$endgroup$
– Kevin Casto
May 31 at 3:31
1
$begingroup$
At first glance it looks right (assuming real scalars), but I also think my answer is right ... going to have to think about this for a minute.
$endgroup$
– Nik Weaver
May 31 at 3:32
2
$begingroup$
@NikWeaver Heuristic answer: $Sym^2(V^* otimes W^*) = Sym^2(V^*) otimes Sym^2(W^*) + Lambda^2(V^*) otimes Lambda^2(W^*)$. This is why, when you fix the inner product on $V$ and $W$, you get something antisymmetric in $V$ and in $W$ (as in my answer). I'm trying to make this precise in terms of kernels vs projections, but in any case counting constants should tell you that (2) doesn't pin down what the total inner product is.
$endgroup$
– Kevin Casto
May 31 at 3:40
2
$begingroup$
Actually, the scalars might be the issue --- my argument for uniqueness uses complex numbers. So we get different answers depending on whether we use real or complex scalars?
$endgroup$
– Nik Weaver
May 31 at 3:41
add a comment |
$begingroup$
Let $V, W$ be 2-dimensional with orthonormal bases $v_0, v_1$ and $w_0, w_1$. If you expand out your condition (2), you obtain the following conditions:
$$ langle v_i otimes w_j, v_i otimes w_j rangle = 1 \
langle v_i otimes w_j, v_i otimes w_k rangle = langle v_i otimes w_j, v_k otimes w_j rangle = 0 \
langle v_0 otimes w_0, v_1 otimes w_1 rangle + langle v_1 otimes w_0, v_0 otimes w_1 rangle = 0
$$
Clearly you can pick any value for $langle v_0 otimes w_0, v_1 otimes w_1 rangle$. Any nonzero choice will not satisfy your condition (1), since the RHS will be 0. So it is certainly possible. I'm not sure if they're of interest.
EDIT: Since there's disagreement in the comments, I'm posting my calculation. Consider general vectors $a_0 v_0 + a_1 v_1 in V$, $b_0 w_0 + b_1 w_1 in W$. Let's mangle Einstein notation by putting $langle v_i otimes w_j, v_k otimes w_lrangle = g_ij^kl$.
(2) says that $ langle a_0 v_0 + a_1 v_1, a_0 v_0 + a_1 v_1 rangle = (a_0^2 + a_1^2)(b_0^2 + b_1^2)$. Expanding out the LHS gives
$$ a_0^2 b_0^2 g_00^00 + a_1^2 b_0^2 g_10^10 + a_0^2 b_1^2 g_01^01 + a_1^2 b_1^2 g_11^11 + \
2left(a_1 a_0 b_0^2 g_10^00 + a_0^2 b_1 b_0 g_01^00 + a_1^2 b_1 b_0 g^10_11 + a_1 a_0 b_1^2 g^01_11right) + \
4a_1 a_0 b_1 b_0 left(g^00_11 + g^10_01 right).$$
By choosing unit vectors, it is clear that $g^ij_ij = 1$. So the first line cancels with the RHS. By choosing a sum of two unit vectors, it is clear that $g^ij_ik = g^ij_kj = 0$, so the second line vanishes. We are left with $4a_1 a_0 b_1 b_0 (g^00_11 + g^10_01)$ = 0. If we choose our $g$ so that $g^00_11 + g^10_01 = 0$, this quantity will always vanish, and so (2) will hold for any pair of vectors.
answered May 31 at 0:52
Kevin CastoKevin Casto
1,21421213
$endgroup$
$begingroup$
This isn't correct; you've only assumed (2) for basis vectors, not all $v in mathcalV$ and $w in mathcalW$.
$endgroup$
– Nik Weaver
May 31 at 3:06
1
$begingroup$
Hi @NikWeaver -- I've posted my calculation, let me know if there's a mistake. Thanks!
$endgroup$
– Kevin Casto
May 31 at 3:31
1
$begingroup$
At first glance it looks right (assuming real scalars), but I also think my answer is right ... going to have to think about this for a minute.
$endgroup$
– Nik Weaver
May 31 at 3:32
2
$begingroup$
@NikWeaver Heuristic answer: $Sym^2(V^* otimes W^*) = Sym^2(V^*) otimes Sym^2(W^*) + Lambda^2(V^*) otimes Lambda^2(W^*)$. This is why, when you fix the inner product on $V$ and $W$, you get something antisymmetric in $V$ and in $W$ (as in my answer). I'm trying to make this precise in terms of kernels vs projections, but in any case counting constants should tell you that (2) doesn't pin down what the total inner product is.
$endgroup$
– Kevin Casto
May 31 at 3:40
2
$begingroup$
Actually, the scalars might be the issue --- my argument for uniqueness uses complex numbers. So we get different answers depending on whether we use real or complex scalars?
$endgroup$
– Nik Weaver
May 31 at 3:41
add a comment |
$begingroup$
Let $V, W$ be 2-dimensional with orthonormal bases $v_0, v_1$ and $w_0, w_1$. If you expand out your condition (2), you obtain the following conditions:
$$ langle v_i otimes w_j, v_i otimes w_j rangle = 1 \
langle v_i otimes w_j, v_i otimes w_k rangle = langle v_i otimes w_j, v_k otimes w_j rangle = 0 \
langle v_0 otimes w_0, v_1 otimes w_1 rangle + langle v_1 otimes w_0, v_0 otimes w_1 rangle = 0
$$
Clearly you can pick any value for $langle v_0 otimes w_0, v_1 otimes w_1 rangle$. Any nonzero choice will not satisfy your condition (1), since the RHS will be 0. So it is certainly possible. I'm not sure if they're of interest.
EDIT: Since there's disagreement in the comments, I'm posting my calculation. Consider general vectors $a_0 v_0 + a_1 v_1 in V$, $b_0 w_0 + b_1 w_1 in W$. Let's mangle Einstein notation by putting $langle v_i otimes w_j, v_k otimes w_lrangle = g_ij^kl$.
(2) says that $ langle a_0 v_0 + a_1 v_1, a_0 v_0 + a_1 v_1 rangle = (a_0^2 + a_1^2)(b_0^2 + b_1^2)$. Expanding out the LHS gives
$$ a_0^2 b_0^2 g_00^00 + a_1^2 b_0^2 g_10^10 + a_0^2 b_1^2 g_01^01 + a_1^2 b_1^2 g_11^11 + \
2left(a_1 a_0 b_0^2 g_10^00 + a_0^2 b_1 b_0 g_01^00 + a_1^2 b_1 b_0 g^10_11 + a_1 a_0 b_1^2 g^01_11right) + \
4a_1 a_0 b_1 b_0 left(g^00_11 + g^10_01 right).$$
By choosing unit vectors, it is clear that $g^ij_ij = 1$. So the first line cancels with the RHS. By choosing a sum of two unit vectors, it is clear that $g^ij_ik = g^ij_kj = 0$, so the second line vanishes. We are left with $4a_1 a_0 b_1 b_0 (g^00_11 + g^10_01)$ = 0. If we choose our $g$ so that $g^00_11 + g^10_01 = 0$, this quantity will always vanish, and so (2) will hold for any pair of vectors.
answered May 31 at 0:52
Kevin CastoKevin Casto
1,21421213
$endgroup$
Let $V, W$ be 2-dimensional with orthonormal bases $v_0, v_1$ and $w_0, w_1$. If you expand out your condition (2), you obtain the following conditions:
$$ langle v_i otimes w_j, v_i otimes w_j rangle = 1 \
langle v_i otimes w_j, v_i otimes w_k rangle = langle v_i otimes w_j, v_k otimes w_j rangle = 0 \
langle v_0 otimes w_0, v_1 otimes w_1 rangle + langle v_1 otimes w_0, v_0 otimes w_1 rangle = 0
$$
Clearly you can pick any value for $langle v_0 otimes w_0, v_1 otimes w_1 rangle$. Any nonzero choice will not satisfy your condition (1), since the RHS will be 0. So it is certainly possible. I'm not sure if they're of interest.
EDIT: Since there's disagreement in the comments, I'm posting my calculation. Consider general vectors $a_0 v_0 + a_1 v_1 in V$, $b_0 w_0 + b_1 w_1 in W$. Let's mangle Einstein notation by putting $langle v_i otimes w_j, v_k otimes w_lrangle = g_ij^kl$.
(2) says that $ langle a_0 v_0 + a_1 v_1, a_0 v_0 + a_1 v_1 rangle = (a_0^2 + a_1^2)(b_0^2 + b_1^2)$. Expanding out the LHS gives
$$ a_0^2 b_0^2 g_00^00 + a_1^2 b_0^2 g_10^10 + a_0^2 b_1^2 g_01^01 + a_1^2 b_1^2 g_11^11 + \
2left(a_1 a_0 b_0^2 g_10^00 + a_0^2 b_1 b_0 g_01^00 + a_1^2 b_1 b_0 g^10_11 + a_1 a_0 b_1^2 g^01_11right) + \
4a_1 a_0 b_1 b_0 left(g^00_11 + g^10_01 right).$$
By choosing unit vectors, it is clear that $g^ij_ij = 1$. So the first line cancels with the RHS. By choosing a sum of two unit vectors, it is clear that $g^ij_ik = g^ij_kj = 0$, so the second line vanishes. We are left with $4a_1 a_0 b_1 b_0 (g^00_11 + g^10_01)$ = 0. If we choose our $g$ so that $g^00_11 + g^10_01 = 0$, this quantity will always vanish, and so (2) will hold for any pair of vectors.
answered May 31 at 0:52
Kevin CastoKevin Casto
1,21421213
edited May 31 at 3:29
answered May 31 at 0:52
Kevin CastoKevin Casto
1,21421213
answered May 31 at 0:52
Kevin CastoKevin Casto
1,21421213
answered May 31 at 0:52
Kevin CastoKevin Casto
1,21421213
1,21421213
$begingroup$
This isn't correct; you've only assumed (2) for basis vectors, not all $v in mathcalV$ and $w in mathcalW$.
$endgroup$
– Nik Weaver
May 31 at 3:06
1
$begingroup$
Hi @NikWeaver -- I've posted my calculation, let me know if there's a mistake. Thanks!
$endgroup$
– Kevin Casto
May 31 at 3:31
1
$begingroup$
At first glance it looks right (assuming real scalars), but I also think my answer is right ... going to have to think about this for a minute.
$endgroup$
– Nik Weaver
May 31 at 3:32
2
$begingroup$
@NikWeaver Heuristic answer: $Sym^2(V^* otimes W^*) = Sym^2(V^*) otimes Sym^2(W^*) + Lambda^2(V^*) otimes Lambda^2(W^*)$. This is why, when you fix the inner product on $V$ and $W$, you get something antisymmetric in $V$ and in $W$ (as in my answer). I'm trying to make this precise in terms of kernels vs projections, but in any case counting constants should tell you that (2) doesn't pin down what the total inner product is.
$endgroup$
– Kevin Casto
May 31 at 3:40
2
$begingroup$
Actually, the scalars might be the issue --- my argument for uniqueness uses complex numbers. So we get different answers depending on whether we use real or complex scalars?
$endgroup$
– Nik Weaver
May 31 at 3:41
add a comment |
$begingroup$
This isn't correct; you've only assumed (2) for basis vectors, not all $v in mathcalV$ and $w in mathcalW$.
$endgroup$
– Nik Weaver
May 31 at 3:06
1
$begingroup$
Hi @NikWeaver -- I've posted my calculation, let me know if there's a mistake. Thanks!
$endgroup$
– Kevin Casto
May 31 at 3:31
1
$begingroup$
At first glance it looks right (assuming real scalars), but I also think my answer is right ... going to have to think about this for a minute.
$endgroup$
– Nik Weaver
May 31 at 3:32
2
$begingroup$
@NikWeaver Heuristic answer: $Sym^2(V^* otimes W^*) = Sym^2(V^*) otimes Sym^2(W^*) + Lambda^2(V^*) otimes Lambda^2(W^*)$. This is why, when you fix the inner product on $V$ and $W$, you get something antisymmetric in $V$ and in $W$ (as in my answer). I'm trying to make this precise in terms of kernels vs projections, but in any case counting constants should tell you that (2) doesn't pin down what the total inner product is.
$endgroup$
– Kevin Casto
May 31 at 3:40
2
$begingroup$
Actually, the scalars might be the issue --- my argument for uniqueness uses complex numbers. So we get different answers depending on whether we use real or complex scalars?
$endgroup$
– Nik Weaver
May 31 at 3:41
$begingroup$
This isn't correct; you've only assumed (2) for basis vectors, not all $v in mathcalV$ and $w in mathcalW$.
$endgroup$
– Nik Weaver
May 31 at 3:06
$begingroup$
This isn't correct; you've only assumed (2) for basis vectors, not all $v in mathcalV$ and $w in mathcalW$.
$endgroup$
– Nik Weaver
May 31 at 3:06
1
1
$begingroup$
Hi @NikWeaver -- I've posted my calculation, let me know if there's a mistake. Thanks!
$endgroup$
– Kevin Casto
May 31 at 3:31
$begingroup$
Hi @NikWeaver -- I've posted my calculation, let me know if there's a mistake. Thanks!
$endgroup$
– Kevin Casto
May 31 at 3:31
1
1
$begingroup$
At first glance it looks right (assuming real scalars), but I also think my answer is right ... going to have to think about this for a minute.
$endgroup$
– Nik Weaver
May 31 at 3:32
$begingroup$
At first glance it looks right (assuming real scalars), but I also think my answer is right ... going to have to think about this for a minute.
$endgroup$
– Nik Weaver
May 31 at 3:32
2
2
$begingroup$
@NikWeaver Heuristic answer: $Sym^2(V^* otimes W^*) = Sym^2(V^*) otimes Sym^2(W^*) + Lambda^2(V^*) otimes Lambda^2(W^*)$. This is why, when you fix the inner product on $V$ and $W$, you get something antisymmetric in $V$ and in $W$ (as in my answer). I'm trying to make this precise in terms of kernels vs projections, but in any case counting constants should tell you that (2) doesn't pin down what the total inner product is.
$endgroup$
– Kevin Casto
May 31 at 3:40
$begingroup$
@NikWeaver Heuristic answer: $Sym^2(V^* otimes W^*) = Sym^2(V^*) otimes Sym^2(W^*) + Lambda^2(V^*) otimes Lambda^2(W^*)$. This is why, when you fix the inner product on $V$ and $W$, you get something antisymmetric in $V$ and in $W$ (as in my answer). I'm trying to make this precise in terms of kernels vs projections, but in any case counting constants should tell you that (2) doesn't pin down what the total inner product is.
$endgroup$
– Kevin Casto
May 31 at 3:40
2
2
$begingroup$
Actually, the scalars might be the issue --- my argument for uniqueness uses complex numbers. So we get different answers depending on whether we use real or complex scalars?
$endgroup$
– Nik Weaver
May 31 at 3:41
$begingroup$
Actually, the scalars might be the issue --- my argument for uniqueness uses complex numbers. So we get different answers depending on whether we use real or complex scalars?
$endgroup$
– Nik Weaver
May 31 at 3:41
add a comment |
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$begingroup$
However, there are various different norms on the tensor products of Hilbert spaces which do not come from an inner product (i.e. they do not satisfy the parallelogram law).
$endgroup$
– Matthias Ludewig
May 31 at 3:59
$begingroup$
There is a construction called the diamond norm that I think goes close to it but maybe not entirely what you want.
$endgroup$
– lcv
May 31 at 4:22
$begingroup$
@lcv It doesn't come from an inner product, and the OP is asking about inner products
$endgroup$
– Yemon Choi
May 31 at 12:03
1
$begingroup$
To be clear - as Yemon Choi says - I am assuming that the norm in (2) is induced by an inner product.
$endgroup$
– Pierre Dubois
May 31 at 12:39
$begingroup$
@PierreDubois no problem yes now it's clearer I can remove my answer.
$endgroup$
– lcv
May 31 at 15:14