# Tag Info

14

I know two-ish answers to this question. Representation-theoretic: The category of representations of a group has both tensor products and duals, but the category of representations of a general algebra generally has neither (or at least there is no obvious way to define them). Since the category of representations of a group $G$ is equivalent to the ...

10

One application of this is that topological K-theory (i.e., the K-theory of the exact category of vector bundles on the space) is the same thing as the "algebraic" $K_0$ of the ring of continuous functions (i.e., the K-theory of the exact category of finitely generated projective modules over that ring). So topological K-theory is a "special case" of ...

8

Roughly, the answer will be that closed C*-subalgebras will correspond to quotient spaces (via pull-back of functions). In your example, the quotient map is one which identifies the two points into a single point. I haven't thought through, though, whether this is a completely correct statement as it stands, or whether one has to add additional caveats. ...

7

You could perhaps do worse than consulting $\S 6.4$ of my commutative algebra notes: "Applications of Swan's Theorem." (You could definitely do better: see below.) The first application I give is to show that the ring of real-valued continuous functions on $[0,1]$ is a connected ring in which each finitely generated projective module is free but for which ...

6

In a recent paper, I adapted some of the terminology of Kochen and Specker's original paper to a more ring-theoretic context. I would refer to your first type of morphism as a morphism of partial $\mathbb{Z}$-algebras from $R$ to $S$, or even better as a morphism of partial rings. The point is that every ring has the underlying structure of a partial ring ...

6

I think Qiaochu's answer is the best, but I thought that two topological reasons should be given. The first is that the Steenrod squares form a Hopf algebra, and their structure is one of the most important aspects of homology theory. The second is the diagrammatic interpretation of the axioms. Letting $\lambda$ denote the multiplication operator, and $Y$ ...

6

Before Drinfeld's work in the 1980s there was only marginal interest by mathematicians in general (noncommutative, noncocommutative) Hopf algebras, so it could be difficult to honestly interest students in the Hopf algebra axioms without mentioning quantum groups. On this view the place to start is Drinfeld's ICM lecture and papers, or books on quantum ...

5

Take $A = L^{\infty}[0,1]$ with pointwise multiplication and let $B = C[0,1]$ be the closed $C^{\ast}$-subalgebra of continuous functions. The $\ast$-subalgebra $D \subset A$ consisting of the simple functions (finite linear combinations of characteristic functions of measurable sets) is dense in $L^{\infty}[0,1]$ but $D \cap B = \{\text{constant ... 5 If I'm reading your question correctly (you may be asking several related things and I am not sure exactly what is a question and what is a discussion), the analogous theorem in the Banach algebra setting is the commutative Gelfand-Naimark theorem. The geometric thing is compact Hausdorff spaces and the algebraic thing is commutative unital C*-algebras. The ... 4 I'm afraid I'm rather late to the party, but let me throw out a few thoughts, in the hope that something will be of use to someone. You probably know everything under 1. and 2., so if you want the punchline, do forgive the tl;dr and just skip ahead to 3. To be absolutely clear about the state of the art, Connes's theorem actually tells you the following: ... 4 You can do this for an arbitrary ring (with or without unit). Jacobson's original article can be found here (JSTOR, needs a university subscription). I cannot do better than to simply quote C. Chevalley's Math Review (MathSciNet, needs a university subscription): A (two-sided) ideal$\frak J$in a ring$\frak A$is called primitive if$0$is the only ... 3 After many days of consistent effort, finally I could find a complete article that speaks about proving the various corollaries and theorems of Gauss-Bonnet , even though the article starts with a preliminary version of the proof ( considering Riemann metrics and the Euler forms, it does have a different versions of the proof, and the whole article is ... 3 One way that Hopf algebras come up is as the algebra of (real or complex) functions on a topological group. The multiplication is commutative since it is just pointwise multiplication of functions. However, in non-commutative geometry you want to replace the algebra of functions on a space with a non-commutative algebra, giving a non-commutative Hopf ... 3 Standard references for basic C*-algebra theory include: C*-Algebras by Example by Kenneth R. Davidson C*-Algebras and Operator Theory by Gerard J. Murphy An Introduction to K-Theory for C*-Algebras by M. Rørdam, F. Larsen, N. Laustsen In general, it seems safe to state that a solid knowledge of algebraic topology and functional analysis is useful for ... 3 It comes from composition of isomorphisms. One version of the "algebra of functions" on, say, a finite groupoid$G$is its groupoid algebra$\mathbb{C}[G]$, which is a direct generalization of the group algebra: take the free vector space on the morphisms in$G$with multiplication given by composition (or$0$if there is no composition). If$G$is an ... 3 1) What is the physical interpretation of q-deformation? It doesn't look like it is the same as going from classical mechanics to quantum mechanics. I have seen q-deformation in the case of quantum groups, though nothing was said about the physical interpretation. A relation between the two notions is If$[x,y]=h$(a scalar or central element) ... 3 I would recommend that you learn some Operator Algbras first, from say Murphy's Operator Algebras and Operator Theory. Then you can learn some K-theory from Rordam/Larsen/Laustsen's book Introduction to K-theory for Operator Algebras. After that, you should be knowledgeable enough to find your own way around. 3 The term you want is "distributes over," not "commutes with." (Whatever "$A$commutes with$B$" means it should be symmetric in$A$and$B$, which the condition you want is not.) Such monoidal categories are called distributive. Examples include any closed monoidal category with coproducts because in this case$A \otimes (-)$has a right adjoint and hence ... 2 I know nothing about noncommutative geometry, but I had wondered this exact thing a while ago and found the following answer. It is part of a report from the BIRS Workshop on Noncommutative Geometry held at the Banff International Research Station in April 2003. The full report is available at www.pims.math.ca/birs. Current applications and connections ... 2 This isn't really an answer to your question (except perhaps the last part), but: Wikipedia claims that the localization of a noncommutative ring$R$with respect to some subset$S$does not always exist. This, I think, comes from using the wrong definition of localization. The definition that seems natural to me is the following. Definition: The ... 2 They are the ($\ell=1$)-spherical harmonics; some help is provided in the reference just after the statement (which is in German): $$a=-\sqrt{\frac{8 \pi}{3}}Y_{1,1} = \mathrm{e}^{\mathrm{i}\phi}\sin \theta, \,\,b=\sqrt{\frac{8 \pi}{3}}Y_{1,-1} = \mathrm{e}^{\mathrm{-i}\phi}\sin \theta,\,\, c:=c_+-c_-=\cos \theta=\sqrt{\frac{4\pi}{3}}Y_{1,0}.$$ where ... 2 One instance where an operator theory problem was solved via "noncommutative topology" (actually it is the introduction of that viewpoint into operator theory) is the Brown-Douglas-Fillmore theory for classifying essentially normal operators. They found that, in addition to the essential spectrum, the Fredholm index was the key ingredient for classification ... 2 I'm not sure what you mean by "tend to contradict each other", but here is one relevant point. An easy way to satisfy 1 would be for$D$to be a bounded operator and$A$to consist of other bounded operators. However, this would make 2 nearly impossible. For if$D$is bounded, then so is$T := (D^2 + 1)^{1/2}$. If$T^{-1}$is compact then it is a ... 2 A survey: http://www.mast.queensu.ca/~speicher/survey.html also worth a read: http://www.ams.org/notices/200405/comm-nas.pdf 1 The basic book is Free Random Variables written by the founder of this theory--D.V.Voiculescu and his followers. It is also helpful to learn the beautiful applications of this theory in von Neumann algebras, especially the free group factors by checking the following survey article--- Ge's report in ICM 2002. Free Probability, Free Entropy and. ... 1 Here is also a more general fact. Let$G$act freely and properly on a space X. Then there is a (in general noncanonical) isomorphism$C_0(X) \rtimes_{r} G \cong C_0(X/G)\otimes \mathcal{K}(\ell^2(G))$which implies nonunitality. Since the reduced crossed product should be a quotient of the full crossed product (is this true?) this also tells about unitality ... 1 I cannot comment, and this should be a comment... Observe that the question in your title and the question in the body of your question are quite different! A non-commutative non-cocommutative Hopf algebra is not the same thing as a non-commutative group, and quantum groups are usually associative. 1 I'm a PhD student in noncommutative algebra - I work on Hopf algebras - and it is definitely, definitely hard to avoid noncommutative geometry. That said, from my vantage point the geometric side of things looks much more algebraic than differential, and the differential sort is less relevant and more easily ignored. It's also... I don't think you need to ... 1 An "ennuple" just sounds like an element of$F^n$, and "right-ratio" just sounds like an equivalence class of ennuples under the equivalence relation$x\sim y$iff$\exists z\in F^\ast$such that$xz=y$. (Here I'm using the$^\ast$to denote the nonzero elements of the ring$F$). If we're talking about a division ring$D$, then an ennuple$x\in D^n\$ would ...

1

As I said in the comment, I'll post an answer, since in the meantime I got to talk to a researcher in the field and somehow made my thoughts clearer. If there aren't any other answers, complaints or relevant comments, I'll accept this in a couple of days and call it a close. My initial thought was that C*-algebras, vector bundles and the like, which are ...

Only top voted, non community-wiki answers of a minimum length are eligible