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  1.  
    • CommentRowNumber1.
    • CommentAuthorTobyBartels
    • CommentTimeAug 22nd 2012

    New page: Banach coalgebra.

    Hopefully you all know that l 1l^1 is a Banach algebra under convolution, but did you know that l ∞l^\infty is a Banach coalgebra under nvolution? (Actually, they are both Banach bialgebras!)

    • CommentRowNumber2.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 22nd 2012

    No, I hadn’t known that – interesting!

    But a part of me hopes that ’nvolution’ won’t become popular! ;-) I thought it was a typo before I realized it was intentional.

    • CommentRowNumber3.
    • CommentAuthorTobyBartels
    • CommentTimeAug 22nd 2012
    • (edited Aug 22nd 2012)

    Although I don’t know another name for it, I’m quite taken with nvolution.

    It’s much simpler than convolution; if ff is any function (taking values in any set) on any abelian group GG (written additively), then the nvolution of ff is a function on G×GG \times G:

    (Δf)(x,y)=f(x+y). (\Delta f)(x,y) = f(x + y) .

    (If GG is nonabelian, the same formula still works, although we also have an anti-nvolution using f(y+x)f(y + x).)

    Then if we fix GG to be locally compact and Hausdorff; then if we also require ff to be complex-valued, continuous, and compactly supported; then we get a Banach coalgebra, whose dual is the Banach algebra of complex-valued Radon measures on GG with convolution. Now I feel like I understand convolution better!

    • CommentRowNumber4.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 22nd 2012

    I need another cocup of ffee. (Hardy-har-har-har)

    • CommentRowNumber5.
    • CommentAuthorTobyBartels
    • CommentTimeAug 22nd 2012

    H’m, it’s more complicated than what I said, since the nvolution of a non-zero compactly supported function is not compactly supported. (The support is an anti-diagonal stripe across G×GG \times G.) So we are back to being lucky.

    • CommentRowNumber6.
    • CommentAuthorMike Shulman
    • CommentTimeAug 22nd 2012

    How do you pronounce “nvolution”? The only way I can think of to pronounce it sounds too much like “involution”.

    • CommentRowNumber7.
    • CommentAuthorTobyBartels
    • CommentTimeAug 23rd 2012

    “nnn-volution” or “en-volution”

    It’s not a good word, I just don’t know another. Well “co-convolution” would be less of a joke, but that’s all.

    • CommentRowNumber8.
    • CommentAuthorzskoda
    • CommentTimeAug 23rd 2012
    • (edited Aug 23rd 2012)

    But there is a cocone so I would be OK with term coconvolution.

    In general, given an kk-algebra (A,m)(A,m) and a kk-coalgebra (C,Δ)(C,\Delta), the convolution product is defined on Hom k(C,A)Hom_k(C,A), : f⋆g=m∘(f⊗g)∘Δf \star g = m\circ (f\otimes g)\circ\Delta. Is there a similar dual picture for coconvolutions in some sense ?

    Another question: coalgebras are simpler than algebras; every coalgebra gives a dual algebra but not otherway around. Then because of finiteness involved one has a coradical filtration and every coalgebra splits into finite pieces. Now this fails for Banach coalgebras. Is there some remnant of in the sense of being simpler in general than Banach algebras ?

    • CommentRowNumber9.
    • CommentAuthorMike Shulman
    • CommentTimeAug 23rd 2012

    I’d prefer coconvolution.

    • CommentRowNumber10.
    • CommentAuthorTobyBartels
    • CommentTimeAug 23rd 2012

    @Zoran: Can convolution on l 1l^1 be made an example of the general set-up in #8? I can’t see the connection.

    (Trivially, if CC is kk, then ⋆\star reduces to mm, so any multiplication is a case of convolution in this sense. But that doesn’t justify using the same term.)

    Is there some remnant of in the sense of being simpler in general than Banach algebras ?

    We do still have the fact that every coalgebra gives rise to an algebra (its dual) but not conversely, but I don’t know what one can conclude from this.

    Judging from Google, very little is written about Banach coalgebras, and much of what is written is not about real or complex Banach spaces (which are otherwise more common) but instead about ultraBanach spaces over nonArchimedean fields. I don’t know why this is. I really don’t know about Banach coalgebras; I’ve just been led to think about them.

    • CommentRowNumber11.
    • CommentAuthorzskoda
    • CommentTimeAug 23rd 2012
    • (edited Aug 23rd 2012)

    Can convolution on l 1l^1 be made an example of the general set-up in #8?

    I think this is one of the basic examples in Majid’s book, in my vague memory. Also in von Neumann context similar examples treated in Enoch-Schwarz.

    • CommentRowNumber12.
    • CommentAuthorTobyBartels
    • CommentTimeAug 23rd 2012

    Majid’s book

    Foundations of Quantum Group Theory or A Quantum Groups Primer? Unfortunately, only the latter is at the library here (although I may be able to get the former through ILL, I don’t know if I’m eligible to use that).

    • CommentRowNumber13.
    • CommentAuthorzskoda
    • CommentTimeAug 23rd 2012
    • (edited Aug 23rd 2012)

    Forget then Majid’s book , convolution for group algebras is treated in Weinstein, da Silva, and I think one should proceed the same way here

    http://math.berkeley.edu/~alanw/Models.pdf from page 73, on

    One should restrict the treatment there to some subspace. But it seems that one has the same structure with a coproduct and a product as I said above.

    • CommentRowNumber14.
    • CommentAuthorTobyBartels
    • CommentTimeAug 24th 2012

    Part of the reason for writing ‘nvolution’ is to get somebody to speak up and say what the proper name is, but we’ll have to go with ‘coconvolution’. But it’s still better to think of convolution as dual to coconvolution, rather than the reverse, since coconvolution is both simpler and more general.

    In Section 11.2, speaking of the space C(G)C(G) of continuous maps on a locally compact group GG, da Silva & Weinstein call coconvolution both ‘comultiplication on C(G)C(G)’ (which is more part of the definition of C(G)C(G) as a bialgebra than a name) and ‘pullback of the multiplication on GG’ (which is more the definition of the operation itself than a name). They don’t seem to ever give it a name as such.

    They do however give some nice symbols: if mm is the multiplication in GG, then coconvolution is m *m^*, while convolution is m *m_*. Here the ** indicates pullback and pushforward, not transpose. (OT: the term in category theory ought to be ‘pullin’, not ‘pullback’.) So what I should have learnt about convolution about measures long ago is simply this: it’s the pushforward of measures by mm (combined with the isomorphism of ℳ(G×G)\mathcal{M}(G \times G) with ℳ(G)⊗^ℳ(G)\mathcal{M}(G) {\displaystyle\hat{\otimes}} \mathcal{M}(G)). But they didn’t tell me that in my measure theory class!

    Anyway, although they talk a lot about algebras and coalgebras (and bialgebras and especially Hopf algebras, which all of these are examples of, even l 1l^1 and l ∞l^\infty), they never seem to discuss the sort of convolution that Zoran mentions in #8. I guess if we take AA to be the ground field in that setup, then ** reduces to Δ *\Delta^* (transpose now), but that seems to be a tenuous connection at best.

    • CommentRowNumber15.
    • CommentAuthorDavidRoberts
    • CommentTimeAug 24th 2012

    But it’s still better to think of convolution as dual to coconvolution

    at least then we have cococonvolution = convolution, which is intuitionistically sound ;-)

    • CommentRowNumber16.
    • CommentAuthorMike Shulman
    • CommentTimeAug 24th 2012

    it’s still better to think of convolution as dual to coconvolution, rather than the reverse, since coconvolution is both simpler and more general.

    Well, it’s not the first pair where the one that got named with a “co” is simpler or more basic. Cohomology comes to mind.

    • CommentRowNumber17.
    • CommentAuthorzskoda
    • CommentTimeAug 24th 2012

    14: Toby, the convolution seems just Δ *\Delta^* from the perspective of the dual Hopf algebra, indeed the group algebra and function algebra are duals and the group algebra has the convolution product while the usual one has the usual product. But, in fact, one first distributes the argument x↦∑ yy⊗y −1xx \mapsto \sum_y y \otimes y^{-1} x, taking this is the coproduct in the sense I listed and then one does some sort of multiplication, which is here usually pairing with respect to some integral. Thus ∫f(y)g(y −1x)\int f(y) g(y^{-1}x). This gives the connection to Fourier transform, canonical WW etc. the perspective from Majid’s book.

    • CommentRowNumber18.
    • CommentAuthorTobyBartels
    • CommentTimeAug 25th 2012

    Now Banach Hopf algebras and Banach **-coalgebras are defined; these should be obvious, but since l 1l^1 and l ∞l^\infty are examples, we might as well mention it.

    • CommentRowNumber19.
    • CommentAuthorTobyBartels
    • CommentTimeAug 26th 2012

    Now C *C^*-coalgebras, but these have their own page and their own discussion thread (where I have a question).

    • CommentRowNumber20.
    • CommentAuthorYemon Choi
    • CommentTimeAug 27th 2012
    • (edited Aug 27th 2012)
    Drive by caveat from one of those pesky analyst (or, as a friend used to say, analysists)

    The claim in 14 above that M(G\times G) is (isomorphic in your favourite sense to) M(G)\hat{\otimes} M(G) is not true, for the usual tensor product you guys put on Ban, unless G is atomic.
    • CommentRowNumber21.
    • CommentAuthorYemon Choi
    • CommentTimeAug 27th 2012
    As Toby has observed in 5, defining comonoids in Ban in the "obvious" way runs into problems where natural examples are ruled out. People who work on C-star-algebraic quantum groups run into similar issues.
    • CommentRowNumber22.
    • CommentAuthorYemon Choi
    • CommentTimeAug 27th 2012
    Have left a comment querying the apparent claim that the proj tp of \ell^\infty with itself is \ell^\infty of the Cartesian product.
    • CommentRowNumber23.
    • CommentAuthorTobyBartels
    • CommentTimeAug 27th 2012

    Yemon, thanks for coming by! (I am still digesting your comments to my question on MathOverflow.)

    The claim in 14 above that M(G×G)M(G\times G) is […] M(G)⊗^M(G)M(G){\displaystyle\hat{\otimes}} M(G) is not true

    That may be a good thing from my perspective! The map M(G)⊗^ πM(G)→M(G×G)M(G) {\displaystyle\hat{\otimes}_\pi} M(G) \to M(G \times G) is obvious (which is what is needed in that comment), but now I can’t see the reverse map. (If there’s a standard counterexample to that, I would like to contemplate it.)

    querying the apparent claim that the proj tp of ℓ ∞\ell^\infty with itself is ℓ ∞\ell^\infty of the Cartesian product

    I’ve formatted the query there, but I’ll answer it here. Again, the map l ∞(ℤ)⊗^ πl ∞(ℤ)l^\infty(\mathbb{Z}) {\displaystyle\hat{\otimes}_\pi} l^\infty(\mathbb{Z}) to l ∞(ℤ×ℤ)l^\infty(\mathbb{Z} \times \mathbb{Z}) is obvious. For the reverse, I claim that any matrix with absolutely bounded entries is the limit1 of its finite submatrices (extended by zeroes), and the latter are obviously in the tensor product. I’m not familiar with Grothendieck’s inequality, but having read its English Wikipedia article, I don’t see how to apply it here.

    1. Specifically, let M≔‖a‖ ∞M \coloneqq {\|a\|_\infty}. Given ε>0\epsilon \gt 0, we have some i,ji,j with M−ε<|a i,j|≤MM - \epsilon \lt {|a_{i,j}|} \leq M. Given any finite submatrix bb including entry (i,j)(i,j), we have M−ε<‖b‖ ∞≤MM - \epsilon \lt {\|b\|_\infty} \leq M, so |M−‖b‖ ∞|=|‖a‖ ∞−‖b‖ ∞|<ε{|M - {\|b\|_\infty}|} = {|{\|a\|_\infty} - {\|b\|_\infty|}} \lt \epsilon. Thus, aa is the limit of the net of finite submatrices. ↩

    • CommentRowNumber24.
    • CommentAuthorTobyBartels
    • CommentTimeAug 27th 2012

    Wait, I guess that I do have a problem! How do I know that the projective tensor norm on l ∞(ℤ×ℤ)l^\infty(\mathbb{Z} \times \mathbb{Z}) (or rather its subspace l ∞(ℤ)⊗l ∞(ℤ)l^\infty(\mathbb{Z}) \otimes l^\infty(\mathbb{Z}), the algebraic tensor product), is the supremum norm?

    It’s obvious that ‖a⊗b‖ ∞=‖a‖‖b‖=‖a⊗b‖ π{\|a \otimes b\|_\infty} = {\|a\|}\, {\|b\|} = {\|a \otimes b\|}_\pi, but I can only get ‖a⊗b+c⊗d‖ ∞≤‖a⊗b‖ ∞+‖c⊗d‖ ∞=‖a‖‖b‖+‖c‖‖d‖≥‖a⊗b+c⊗d‖ π{\|a \otimes b + c \otimes d\|_\infty} \leq {\|a \otimes b\|_\infty} + {\|c \otimes d\|_\infty} = {\|a\|}\, {\|b\|} + {\|c\|}\, {\|d\|} \geq {\|a \otimes b + c \otimes d\|_\pi} so far. I find that I am not very good at figuring out the different ways to write even a 2×22 \times 2 matrix as a sum of tensor products of 22-tuples, so this is getting difficult.

    Is this your concern? (I still don’t see how Grothendieck’s inequality helps.)

    • CommentRowNumber25.
    • CommentAuthorYemon Choi
    • CommentTimeAug 27th 2012

    Hi Toby. Still a bit too busy to engage properly - apologies for any perceived curtness, I was typing last night in a rush.

    You say

    How do I know that the projective tensor norm on l ∞(ℤ×ℤ)l^\infty(\mathbb{Z} \times \mathbb{Z}) (or rather its subspace l ∞(ℤ)⊗l ∞(ℤ)l^\infty(\mathbb{Z}) \otimes l^\infty(\mathbb{Z}), the algebraic tensor product), is the supremum norm?

    Well, you can’t, because it isn’t; not even up to equivalence of norms. I don’t have the references to hand where I first learned this, I’m afraid. The reason I mentioned Grothendieck’s inequality is because one form of it tells us things about the projective tensor norm on ℓ ∞⊗ℓ ∞\ell^\infty\otimes\ell^\infty and these are incompatible with it being the supremum norm in the sense you describe. E *⊗^ πF *E^*\hat{\otimes}_\pi F^* maps naturally to (E⊗ πF) *(E\otimes_\pi F)^* but in general this map does not have closed range (and in particular it can’t be an isometry).

    In fact, I am fairly sure that there is no TVS isomorphism from ℓ ∞⊗ πℓ ∞\ell^\infty\otimes_\pi \ell^\infty to ℓ ∞\ell^\infty, but I think all known proofs require some heavy duty Banach space machinery.

    What you do have is C(K 1)⊗ εC(K 2)C(K_1)\otimes_\varepsilon C(K_2) naturally (isometrically) isomorphic to C(K 1×K 2)C(K_1\times K_2) where ⊗ ε\otimes_\varepsilon denotes the so-called injective tensor norm of Banach spaces. The quickest way to define said norm is that it is the norm you get on E⊗FE\otimes F by naturally embedding said space into Hom rmBan(E *,F)\operatorname{Hom}_{\rm Ban}(E^*,F).

    • CommentRowNumber26.
    • CommentAuthorTobyBartels
    • CommentTimeAug 28th 2012

    apologies for any perceived curtness

    Not a problem!

    you can’t, because it isn’t

    Then I can stop trying. (^_^)

    E *⊗^ πF *E^*\hat{\otimes}_\pi F^* maps naturally to (E⊗ πF) *(E\otimes_\pi F)^* but in general this map does not have closed range (and in particular it can’t be an isometry).

    When you put it that way, my claim seems quite unreasonable; I would not expect (V *⊗W *)≅(V⊗W) *(V^* \otimes W^*) \cong (V \otimes W)^* normally, so why expect it for l 1l^1?

    In fact, it just now occurs to me that of course for l 2l^2, we need the tensor product as Hilbert spaces to get l 2(ℤ)⊗^ σl 2(ℤ)≅l 2(ℤ×ℤ)l^2(\mathbb{Z}) {\displaystyle\hat{\otimes}_\sigma} l^2(\mathbb{Z}) \cong l^2(\mathbb{Z} \times \mathbb{Z}), so clearly I can’t expect the projective tensor product to work across the board. It’s starting to look like a very l 1l^1 sort of tensor product (much like the l 1l^1-direct sum).

    By the way, when I search for phrases like ‘Banach coalgebra’, ‘Banach Hopf algebra’, etc; I mostly get results for nonArchimedean Banach spaces. These use the projective tensor product, so I figured that this was the way to go. (Upon reflection, there is no clear alternative here; since the Hahn–Banach Theorem fails for nonArchimedean spaces, the injective tensor product doesn’t really make sense.)

    So if you know references for Banach coalgebras (possibly with some extra structure or properties, such as Hopf or C *C^*) using other tensor products, then I would like to read them.

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