Mathematics Start‑class Low‑priority

This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.MathematicsWikipedia:WikiProject MathematicsTemplate:WikiProject Mathematicsmathematics articles Start This article has been rated as Start-class on Wikipedia's content assessment scale.  Low  This article has been rated as Low-priority on the project's priority scale.

As far as I can tell, the transpose ^tf : W^∗ → V^∗ of a linear map f : V → W is defined independently of any bilinear forms B_V : V × V → K and B_W : W × W → K, yet the second part of § Definition uses a duality pairing that (by the implication of the recently added link) relies on the existence of such bilinear forms. While they can presumably be related if defined and adhering to suitable constraints (and may be a confusion with the adjoint), this seems completely superfluous to the definition an understanding of the transpose defined here. Or perhaps the duality pairing is simply a map V^∗ × V → K (the action of the covector on a vector), in which case the link should be revised. —Quondum 17:53, 4 February 2015 (UTC)Reply[reply]

Hi, @Quondum:I am very sorry for having done something possibly beyond my competence. I was looking for "transpose" and encountered "duality pairing" which I had to look up outside of Wikipedia (Paul Garrett, Abstract Algebra), and which was connected there to BLFs, and so I inserted this link, which you consider to be undue. I hope, the rest I did is a correct edit of a typo and and a more foolproof formulation. I'm going to revert the inserted link. True for finite dimension? Purgy (talk) 20:18, 4 February 2015 (UTC)Reply[reply]

I agree with your initial impulse to link it, and I've found a suitable definition that I've linked to. The word transpose gets used to mean so many similar but nonequivalent things that this gets confusing. I think that it may help to additionally define the duality pairing explicitly here, e.g. [w,v] = w(v), the defined action of the dual vector w on the vector v. What do you think? I do not know enough about the infinite-dimensional case, but it would seem to apply there too. —Quondum 22:56, 4 February 2015 (UTC)Reply[reply]

Definitely, your's is the appropriate link. I think it suffices, and there is no need for a rewriting of the given brackets with new variable names to explain the involved bilinear maps. Checking the links you and I gave, I noticed the tar pit I jumped in, blinded by the simple occurrance of "dual space" in the link: bilinear map allows for two arbitrary spaces, whereas bilinear form restricts itself to twice one space only in mapping to the relevant field. Thanks for the communication and sorry again for my flippancy. Purgy (talk) 14:59, 5 February 2015 (UTC)Reply[reply]

Relax – your engagement has entirely constructive throughout, being bold is encouraged and I really don't see what there is to apologize about. And I'm not particularly sure-footed here, just interested in details.

I'm still interested in exploring the use of the [·,·]-notation in this case. I imagine that it is usually used to emphasize the symmetry between a space and its dual through the identification of a space with its double dual, but that does not seem to be the point here. Since the notation is also uncommon, the question arises of whether it should be removed entirely instead of using it but having to explain it. The duality pairing is nothing but the application operator, and this should be expressible with the notation already employed. I have a feeling that the expression follows rather directly from the definition preceding it, and is thus not saying much. —Quondum 17:08, 5 February 2015 (UTC)Reply[reply]

Thanks for your kind words! I'll stop apologizing and will express my personal dissatisfaction with edits of mine in an other form in the future. :)

I have seen a lot of notation around this already and I do not really appreciate one kind especially. The problem of denoting what acts on what is delt with in many ways. Here its is the brackets. I'm not sure if ${\displaystyle f^{*}(\varphi )(v)=\varphi (f(v))}$ or ${\displaystyle (f^{*}(\varphi ))(v)=\varphi fv}$ is any better. I remember physicists defining scalar products as a map V^∗ x V → R to save one explicit level of mapping. Perhaps it's just a matter of accomodation? Purgy (talk) 10:34, 7 February 2015 (UTC)Reply[reply]

You've expressed my direction of thought exactly. I was mentally trying to picture how understandable it would be in the form you've written it, and concur that it doesn't help. I guess the notation in general (overloading of parentheses etc.) is what I find awkward, but I'm stuck with that. So I think we leave it as is. —Quondum 16:39, 7 February 2015 (UTC)Reply[reply]

I was looking at this entry and comparing it to Raising and lowering indices. I think it would be worth mentioning the connection here. Given a nonsingular bilinear or hermitian product, one can identify the vector space with its dual, which is the simplest example of "lowering indices". The transpose of a linear operator on a vector space can then be defined as a linear operator on the same vector space rather than on its dual. — Preceding unsigned comment added by 2A02:C7D:5E4F:E900:AC6B:ECC0:792A:82DC (talk) 11:56, 24 March 2019 (UTC)Reply[reply]