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5

Howard Georgi's "Lie Algebras in Particle Physics" is good, if more intended for the physicist going towards the math than vice versa. It should provide a lot of context, though, and there's a PDF version floating around on google. I'd say similar things about these two introductions to aspects of high-energy theory [1] [2]. I'll see if I can remember some ...


3

Peter Woit, the author of the book "Not Even Wrong" and a blog by the same name, has been working on a book on quantum mechanics as described by representation theory. The latest draft may be found at the following link: Quantum Theory, Groups and Representations: An Introduction.


3

The free module $RG$ is certainly projective, and there is a surjective $RG$-module homomorphism $\varphi:RG\to R$ where $$\varphi\left(\sum_{g\in G}a_gg\right)=\sum_{g\in G}a_g,$$ so $R$ is projective if and only if $\varphi$ splits. By considering possible splitting maps, it's easy to see that this is the case if and only if $G$ is finite and $|G|$ is a ...


1

You seem to worry about these sentences: "Let $\lambda : G \to S_G$ be the permutation representation afforded by the corresponding right action of $G$ on itself, and for each $h \in G$ denote the permutation $\lambda(h)$ by $\tau_h$. Thus $\tau_h(x)=xh^{−1}$ for all $x \in G$ ($\lambda$ is called the right regular representation of $G$)." What ...


1

It is not true that a (finite-dimensional, complex) representation of $\mathfrak{su}(3)$ is determined by its dimension. In fact it's already the case, as James Cook points out in the comments, that the defining representation $V$ (which I think is the representation you denote by $\mathbf{3}$) and its dual $V^{\ast}$ are not isomorphic. One way to see this ...


1

Are there some references about this fact? Yes, the best references are: V.G. Drinfeld, Quantum Groups, in Proc. of the Int. Conf. of Mathematicians (Berkeley, 1986) V.G. Drinfeld, Sov. Math. Dokl.28 (1983) 667 M. Jimbo, Lett. Math. Phys. 10 (1985) 63, ibid. 11 (1986) 247 M. Jimbo, Commun. Math. Phys. 102 (1986) 537 E. K. Sklyanin, Funct. Anal. Appl. ...


1

Come on, this is really not difficult. The left regular representation of $c$ is the matrix of the left translation by $c$ in the algebra $\mathbb{Z}S_3$. Thus, according to your table: $ca=f$, $cb=d$, $cc=e$, $cd=b$, $ce=c$, $cf=a$. In other words, $ca=0a + 0b + 0c + 0d + 0e + 1f$, $cb=0a + 0b + 0c + 1d + 0e + 0e$, $cc=0a + 0b + 0c + 0d + 1e + 0f$, ...



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