This is to expand on my earlier comment. It really deserves to be part its own post, and in fact I was already in the process of writing that post when I came across Jessica’s work here. Her explanation for the antilinear-linear inner product is interesting—and initially seemed like it was doing something right—but completely different from my own approach. Well actually, I didn’t have an explanation for why the inner product was antilinear-linear, I had only gotten to the point where I knew there was a bilinear form for physical observables, so I was hoping to incorporate some of Jessica’s ideas into my own post. However, there was one detail she was missing, and that was, “why is it a bilinear form in the first place?” I had the explanation, but I could not see it materializing from her approach. Ultimately, I’ve concluded that the reason her approach seems to work is coincidental, and the mirror, conjugate space is not related to the inner product.
So, why a bilinear form? It is because observations are gauge-invariant, and gauge-invariant things are generated by the field strength tensor F=D∧D, where D=d+A is the contravariant derivative along the spacetime manifold. Why does F=∧2D rather than D or ∧3D? Because d2=0.
The boundary or derivative operator d comes from simplicial complexes:
d[v1∧v2∧⋯∧vn]=n∑i=1sgn(12…i)[v1∧⋯∧vi−1∧vi+1∧⋯∧vn]
These simplicial compelexes are in turn one piece of the tensor product. For example:
v1⊗v2=12![(v1⊗v2+v2⊗v1)+(v1⊗v2−v2⊗v1)]
We can identify
v1∧v2=v1⊗v2−v2⊗v1
which matches
sgn(σ)v1∧v2=vσ1∧vσ2.
It seems rather strange to do this, until you recall the Schur-Weyl duality. Consider the tensor space (Cn)⊗n and group actions
These group actions commute, so they are mutual centralizers and decompose into irreducible representations together. We can decompose tensors into a direct sum of these irreducible representations, i.e. break them up into their “symmetric pieces”:
(Cn)⊗n=⨁λ⊢nλ′1≤nSλ⊗Vλ
The corresponding group actions for each piece are represented by πλ⊗ρλ. For the antisymmetric, or sgn representation, Sλ⊗Vλ is a one-dimensional vector space—it contains one basis vector, v1∧v2∧⋯∧vn, and multiples of that vector. This means sgn consists of 1×1 matrices (or scalars), and has lots of nice properties due to being abelian. Rotations will be smooth and transitive, as speck1447 explains here. This is why the boundary operator typically uses this irreducible representation.
However, we can define a more general boundary d as a map to a symmetric piece of one degree lower
d:Vλ⊗Sλ→Vμ⊗Sλ
where μ⊂λ and |μ|=|λ|−1. It looks exactly the same, but we use πλ⊗ρλ rather than sgn to permute an index to the front and drop it. The reason d2=0 for the sgn representation is because there are 2! permutations you can drop two indices in, and summing the different permutations gives you
∑σ∈S2sgn(σ)=0.
In general, you will get dk=0 when
∑σ∈Skπλ(σ)=0.
The minimal k is not always two, so other representations will ultimately lead to multilinear forms of different degrees. An equivalent way to look at it is from the GL(n,C) side. Rotations are essentially continuous permutations, so the generalized idea of a rotation is an n×n matrix A where
ρλ(A)=IdimVλ
(the kernel of ρλ). Mirroring and other orthogonal matrices are generalized to
ρλ(A)∈Im(πλ).
We are looking for a multilinear form B:(Vλ)⊗k→C that is invariant to orthogonal transformations, so for any v∈(Vλ)⊗k and g∈Im(πλ)
B(gv)=B(v)
(note: g is applied diagonally). It is enough to find a single vector where gv=v for all g∈Im(πλ). Then we can set B(v)=1 and for non-multiples of v, B(w)=0. The number of antisymmetrizers is the number of rows λ′1, so there will always be some g where gv=−v≠v until k>λ′1. This gives the minimal k=λ′1+1.
This equivalent view is closer to what Jessica is doing. You can think of (Vλ)⊗k as the original vector space plus several “dual spaces”. However, the “mirror images” come from Im(πλ), and
In terms of goals of OP, I was taking standard QM notation at face value, and trying to peel back one layer of the onion: why antilinear-linear inner product, POVM observables as Hermitians implying quadratic forms, quadratic density matrices, etc? And I think real structure + conjugate spaces basically explains these dualities in the Hilbert space standard formulation, although that doesn’t mean this is the deepest symmetry layer (it’s not).
field strength tensor
This is not an area I understand well, but, either the field values are expressed as a real number or complex number. If real: OP framework doesn’t apply, or applies at a lower level; perhaps applies indirectly through CPT. If complex: Then I’m guessing similar principles will apply, where for example you can “make a complex number real” as ¯¯¯λλ=|λ|2, yielding an inner product that looks like the Hilbert space inner product. The thing to look for would be, are complex conjugates showing up all over the place in the standard math? If so, it may perhaps be modeled with real structure & complex conjugate spaces.
we should not expect to be able to find dual spaces by looking at mirror images
For Hilbert space I’m applying Riesz representation for that; standard machinery for bra-ket. Not sure how far the principle generalizes. (Much of OP could work without Riesz representation, e.g. we can still define A⊗(¯¯¯u⊗v)=⟨¯¯¯u⊗Av⟩ just fine, it’s just that (−)⊗ might not be an invertible map.)
We can identify v1∧v2=v1⊗v2−v2⊗v1
This is looking a bit like the ‘swap’ real structure on ¯¯¯¯V⊗V. The main difference is that it’s dealing with real rather than complex numbers. With the swap real structure, the ‘imaginary component’ would be written as (¯¯¯u⊗v)↦12(¯¯¯¯¯¯¯¯¯¯¯¯u−v⊗v−u). This of course looks similar to u∧v.
Here’s one place where the analogy with symmetric group permutations breaks down. Say our Hilbert space is H=H1⊗H2. Now our inner product (written to be bilinear) is ⟨¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯a1⊗a2,b1⊗b2⟩=⟨¯¯¯¯¯a1,b1⟩⟨¯¯¯¯¯a2,b2⟩. We can see this as quadri-linear in ¯¯¯¯¯a1,¯¯¯¯¯a2,b1,b2. Notice with ‘n-linearity’ we always have n even, and the elements come in pairs. We pair ¯¯¯¯¯a1 with b1 and ¯¯¯¯¯a2 with b2.
What the standard Hilbert space framework is going to do is, by bundling H1⊗H2, make the H1 and H2 bilinearity explicit, with the H1 and ¯¯¯¯¯¯¯H1 bilinearity implicit (through the inner product and bra/kets and so on). This is why we end up with quadratics all over the formalism: Born rule, inner product, density matrices, POVMs.
So for your analogy to hold, it’s critical that n be even. And what the ‘swap’ real structure on ¯¯¯¯¯H⊗H is going to do is swap elements with their conjugated elements (¯¯¯¯¯¯¯H1 with H1, and ¯¯¯¯¯¯¯H2 with H2). This is going to be a specific involutive permutation.
So I guess if you wanted to get into deeper foundations, then a question to ask would be “why do the tensor products that POVM observables are linear in factor into components that come in pairs with their complex conjugate spaces (or something iso to their complex conjugate spaces)?”. Which, again, OP isn’t very much about.
Perhaps I’m missing something you’re saying, or you’re missing something I’m saying. In general, the process of finding a “conjugate space” is not an involution. We do not have inner products or Hilbert spaces. There are no pairs. We have to motivate the inner product, and it is motivated by first motivating multilinear forms, and then motivating bilinear forms. But bilinear forms only arise in the alternating representation.
There is a terminology issue here, because “dual” literally means a paired, mirror space, so treating the inner product as a bilinear form in a space ⊗ its dual only works in the alternating representation. What you’re actually doing to generate that dual space is to look at the connected components of
{A:ρλ(A)∈Im(πλ)}.
For the alternating representation, there is the component connected to the origin—SO(n)—as well as the mirrored component J−1SO(n)J. But there are many more connected components for other representations, exactly |Im(πλ)| of them. Maybe some of them have involutions to each other, but not all of them. Not all of them are “dual” or “conjugate” in the literal sense of the word. This is the terminology issue, and I think the main source of confusion.
Also, my analogy does not break down for odd n. Note that Vλ is an irreducible module of (Cn)⊗n, not the same space, and (Vλ)⊗k is an entirely new tensor product. I was trying to keep my comment from growing longer than it already was, so I may have left out many other little details like this that would help with interpretation.
To disambiguate terminology: By “dual space” I mean the standard meaning in QM; we go from a complex vector space V to a space V∨ of linear maps V→C, with the bra/ket duality as a special case. Perhaps I should avoid using “dual” but this explains my earlier usage.
By “complex conjugate space” (sometimes abbreviated as “conjugate space”) I mean specifically the ¯¯¯¯V construction, a completely formal mathematical operation. (To avoid confusion I could say “complex conjugate space”)
Since “complex conjugate space” is an entirely formal construction, it doesn’t necessarily have a physical meaning. Or it could have multiple physical meanings as different isos with V.
OP doesn’t try to go lower-level than Hilbert space; what follows is my attempt to engage with the level you’re talking about:
A “duality” like thing is “polarity of representation of U(1)±” as implied by the category [U(1)±,VectR]. Where to simplify in OP, I’m always using the polarity between V and ¯¯¯¯V in the representation, but this is not strictly necessary.
The correspondence with the SO(n) model might be: When you are representing an element of [BSO(n),VectR] as a complex vector space (B means “delooping groupoid”), you are picking out a sub-group of SO(n) iso to U(1)≅SO(2). To get to a Hilbert space you clearly have to at least pick out a U(1) subgroup.
Another idea is to find a functor [U(1)±,BO(n)]. This does not necessarily get you an O(2) subgroup, because J and J−1 in U(1)± need not map to the same group element of O(n). (Hence spinors and so on)
If you have a representation [BO(n),VectR] and a groupoid functor [U(1)±,BO(n)] you can trivially compose to get a polar representation [U(1)±,VectR]. That gives you, as the positive component, something like a Hilbert space, and as the negative component, something formally iso to its complex conjugate space (though with a better physical interpretation).
At this point you can use the formal isos: ¯¯¯¯V≅V∨ (Riesz representation), and the iso from ¯¯¯¯V to the polar negative of your Hilbert-like space (where polar negation is from the [U(1)±,VectR] representation). This gives you nice physical interpretations of bra/kets, density matrices, observables, and so on, through formal isos.
(I am not sure how much I’m understanding or how much is connecting; feel free to ignore irrelevant detail)
This is to expand on my earlier comment. It really deserves to be part its own post, and in fact I was already in the process of writing that post when I came across Jessica’s work here. Her explanation for the antilinear-linear inner product is interesting—and initially seemed like it was doing something right—but completely different from my own approach. Well actually, I didn’t have an explanation for why the inner product was antilinear-linear, I had only gotten to the point where I knew there was a bilinear form for physical observables, so I was hoping to incorporate some of Jessica’s ideas into my own post. However, there was one detail she was missing, and that was, “why is it a bilinear form in the first place?” I had the explanation, but I could not see it materializing from her approach. Ultimately, I’ve concluded that the reason her approach seems to work is coincidental, and the mirror, conjugate space is not related to the inner product.
So, why a bilinear form? It is because observations are gauge-invariant, and gauge-invariant things are generated by the field strength tensor F=D∧D, where D=d+A is the contravariant derivative along the spacetime manifold. Why does F=∧2D rather than D or ∧3D? Because d2=0.
The boundary or derivative operator d comes from simplicial complexes:
d[v1∧v2∧⋯∧vn]=n∑i=1sgn(12…i)[v1∧⋯∧vi−1∧vi+1∧⋯∧vn]These simplicial compelexes are in turn one piece of the tensor product. For example:
v1⊗v2=12![(v1⊗v2+v2⊗v1)+(v1⊗v2−v2⊗v1)]We can identify
v1∧v2=v1⊗v2−v2⊗v1which matches
sgn(σ)v1∧v2=vσ1∧vσ2.It seems rather strange to do this, until you recall the Schur-Weyl duality. Consider the tensor space (Cn)⊗n and group actions
σ(v1⊗v2⊗⋯⊗vn)=vσ1⊗vσ2⊗⋯⊗vσnσ∈Sng(v1⊗v2⊗⋯⊗vn)=(gv1)⊗(gv2)⊗⋯⊗(gvn)g∈GL(n,C)These group actions commute, so they are mutual centralizers and decompose into irreducible representations together. We can decompose tensors into a direct sum of these irreducible representations, i.e. break them up into their “symmetric pieces”:
(Cn)⊗n=⨁λ⊢nλ′1≤nSλ⊗VλThe corresponding group actions for each piece are represented by πλ⊗ρλ. For the antisymmetric, or sgn representation, Sλ⊗Vλ is a one-dimensional vector space—it contains one basis vector, v1∧v2∧⋯∧vn, and multiples of that vector. This means sgn consists of 1×1 matrices (or scalars), and has lots of nice properties due to being abelian. Rotations will be smooth and transitive, as speck1447 explains here. This is why the boundary operator typically uses this irreducible representation.
However, we can define a more general boundary d as a map to a symmetric piece of one degree lower
d:Vλ⊗Sλ→Vμ⊗Sλwhere μ⊂λ and |μ|=|λ|−1. It looks exactly the same, but we use πλ⊗ρλ rather than sgn to permute an index to the front and drop it. The reason d2=0 for the sgn representation is because there are 2! permutations you can drop two indices in, and summing the different permutations gives you
∑σ∈S2sgn(σ)=0.In general, you will get dk=0 when
∑σ∈Skπλ(σ)=0.The minimal k is not always two, so other representations will ultimately lead to multilinear forms of different degrees. An equivalent way to look at it is from the GL(n,C) side. Rotations are essentially continuous permutations, so the generalized idea of a rotation is an n×n matrix A where
ρλ(A)=IdimVλ(the kernel of ρλ). Mirroring and other orthogonal matrices are generalized to
ρλ(A)∈Im(πλ).We are looking for a multilinear form B:(Vλ)⊗k→C that is invariant to orthogonal transformations, so for any v∈(Vλ)⊗k and g∈Im(πλ)
B(gv)=B(v)(note: g is applied diagonally). It is enough to find a single vector where gv=v for all g∈Im(πλ). Then we can set B(v)=1 and for non-multiples of v, B(w)=0. The number of antisymmetrizers is the number of rows λ′1, so there will always be some g where gv=−v≠v until k>λ′1. This gives the minimal k=λ′1+1.
This equivalent view is closer to what Jessica is doing. You can think of (Vλ)⊗k as the original vector space plus several “dual spaces”. However, the “mirror images” come from Im(πλ), and
|Im(πλ)|=|Sn/ker(πλ)|=⎧⎨⎩1λ=(n)2λ=(1,1,…,1)n!otherwise.So, for any non-trivial and non-alternating representation,
λ′1+1≤n<n!There are way more mirror images than dual spaces! It is almost a lucky coincidence that
λ′1+1=2=|Im(πλ)|for the alternating representation, and in general, we should not expect to be able to find dual spaces by looking at mirror images.
In terms of goals of OP, I was taking standard QM notation at face value, and trying to peel back one layer of the onion: why antilinear-linear inner product, POVM observables as Hermitians implying quadratic forms, quadratic density matrices, etc? And I think real structure + conjugate spaces basically explains these dualities in the Hilbert space standard formulation, although that doesn’t mean this is the deepest symmetry layer (it’s not).
This is not an area I understand well, but, either the field values are expressed as a real number or complex number. If real: OP framework doesn’t apply, or applies at a lower level; perhaps applies indirectly through CPT. If complex: Then I’m guessing similar principles will apply, where for example you can “make a complex number real” as ¯¯¯λλ=|λ|2, yielding an inner product that looks like the Hilbert space inner product. The thing to look for would be, are complex conjugates showing up all over the place in the standard math? If so, it may perhaps be modeled with real structure & complex conjugate spaces.
For Hilbert space I’m applying Riesz representation for that; standard machinery for bra-ket. Not sure how far the principle generalizes. (Much of OP could work without Riesz representation, e.g. we can still define A⊗(¯¯¯u⊗v)=⟨¯¯¯u⊗Av⟩ just fine, it’s just that (−)⊗ might not be an invertible map.)
This is looking a bit like the ‘swap’ real structure on ¯¯¯¯V⊗V. The main difference is that it’s dealing with real rather than complex numbers. With the swap real structure, the ‘imaginary component’ would be written as (¯¯¯u⊗v)↦12(¯¯¯¯¯¯¯¯¯¯¯¯u−v⊗v−u). This of course looks similar to u∧v.
Here’s one place where the analogy with symmetric group permutations breaks down. Say our Hilbert space is H=H1⊗H2. Now our inner product (written to be bilinear) is ⟨¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯a1⊗a2,b1⊗b2⟩=⟨¯¯¯¯¯a1,b1⟩⟨¯¯¯¯¯a2,b2⟩. We can see this as quadri-linear in ¯¯¯¯¯a1,¯¯¯¯¯a2,b1,b2. Notice with ‘n-linearity’ we always have n even, and the elements come in pairs. We pair ¯¯¯¯¯a1 with b1 and ¯¯¯¯¯a2 with b2.
What the standard Hilbert space framework is going to do is, by bundling H1⊗H2, make the H1 and H2 bilinearity explicit, with the H1 and ¯¯¯¯¯¯¯H1 bilinearity implicit (through the inner product and bra/kets and so on). This is why we end up with quadratics all over the formalism: Born rule, inner product, density matrices, POVMs.
So for your analogy to hold, it’s critical that n be even. And what the ‘swap’ real structure on ¯¯¯¯¯H⊗H is going to do is swap elements with their conjugated elements (¯¯¯¯¯¯¯H1 with H1, and ¯¯¯¯¯¯¯H2 with H2). This is going to be a specific involutive permutation.
So I guess if you wanted to get into deeper foundations, then a question to ask would be “why do the tensor products that POVM observables are linear in factor into components that come in pairs with their complex conjugate spaces (or something iso to their complex conjugate spaces)?”. Which, again, OP isn’t very much about.
Perhaps I’m missing something you’re saying, or you’re missing something I’m saying. In general, the process of finding a “conjugate space” is not an involution. We do not have inner products or Hilbert spaces. There are no pairs. We have to motivate the inner product, and it is motivated by first motivating multilinear forms, and then motivating bilinear forms. But bilinear forms only arise in the alternating representation.
There is a terminology issue here, because “dual” literally means a paired, mirror space, so treating the inner product as a bilinear form in a space ⊗ its dual only works in the alternating representation. What you’re actually doing to generate that dual space is to look at the connected components of
{A:ρλ(A)∈Im(πλ)}.
For the alternating representation, there is the component connected to the origin—SO(n)—as well as the mirrored component J−1SO(n)J. But there are many more connected components for other representations, exactly |Im(πλ)| of them. Maybe some of them have involutions to each other, but not all of them. Not all of them are “dual” or “conjugate” in the literal sense of the word. This is the terminology issue, and I think the main source of confusion.
Also, my analogy does not break down for odd n. Note that Vλ is an irreducible module of (Cn)⊗n, not the same space, and (Vλ)⊗k is an entirely new tensor product. I was trying to keep my comment from growing longer than it already was, so I may have left out many other little details like this that would help with interpretation.
To disambiguate terminology: By “dual space” I mean the standard meaning in QM; we go from a complex vector space V to a space V∨ of linear maps V→C, with the bra/ket duality as a special case. Perhaps I should avoid using “dual” but this explains my earlier usage.
By “complex conjugate space” (sometimes abbreviated as “conjugate space”) I mean specifically the ¯¯¯¯V construction, a completely formal mathematical operation. (To avoid confusion I could say “complex conjugate space”)
Since “complex conjugate space” is an entirely formal construction, it doesn’t necessarily have a physical meaning. Or it could have multiple physical meanings as different isos with V.
OP doesn’t try to go lower-level than Hilbert space; what follows is my attempt to engage with the level you’re talking about:
A “duality” like thing is “polarity of representation of U(1)±” as implied by the category [U(1)±,VectR]. Where to simplify in OP, I’m always using the polarity between V and ¯¯¯¯V in the representation, but this is not strictly necessary.
The correspondence with the SO(n) model might be: When you are representing an element of [BSO(n),VectR] as a complex vector space (B means “delooping groupoid”), you are picking out a sub-group of SO(n) iso to U(1)≅SO(2). To get to a Hilbert space you clearly have to at least pick out a U(1) subgroup.
Another idea is to find a functor [U(1)±,BO(n)]. This does not necessarily get you an O(2) subgroup, because J and J−1 in U(1)± need not map to the same group element of O(n). (Hence spinors and so on)
If you have a representation [BO(n),VectR] and a groupoid functor [U(1)±,BO(n)] you can trivially compose to get a polar representation [U(1)±,VectR]. That gives you, as the positive component, something like a Hilbert space, and as the negative component, something formally iso to its complex conjugate space (though with a better physical interpretation).
At this point you can use the formal isos: ¯¯¯¯V≅V∨ (Riesz representation), and the iso from ¯¯¯¯V to the polar negative of your Hilbert-like space (where polar negation is from the [U(1)±,VectR] representation). This gives you nice physical interpretations of bra/kets, density matrices, observables, and so on, through formal isos.
(I am not sure how much I’m understanding or how much is connecting; feel free to ignore irrelevant detail)