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In Howard Georgi's Lie Algebras in Particle Physics, 2nd Edition, p. 26-onwards, the author of that book considers a system of three equal masses linked together by identical springs as an illustrative example of tensor products and product representations.

However, something he writes greatly confuses me. From the aforementioned work:

"Three blacks are connected by springs in a triangle [...] Suppose that these are free to slide on a frictionless surface. What can we say about the normal modes of this system. The point is that these is an $S_3$ symmetry of the system, and we can learn a lot about the system by using the symmetry and applying theorem 1.6. The system has 6 degrees of freedom, described by the $x$ and $y$ coordinates of the three blocks: \begin{equation} \begin{pmatrix} x_1 & y_1 & x_2 & y_2 & x_3 & y_3 \end{pmatrix} \end{equation} This has the structure of a tensor product $-$ the 6 dimensional space is a product of a 3 dimensional space of the blocks, and the 2 dimensional space of the $x$ and $y$ coordinates."

From this, Georgi seems to draw the conclusion that the representation of the normal modes must be a tensor product of a 3 dimensional representation and a 2 dimensional representation.

I fail to see how this follows. Surely Georgi cannot be saying that if the solution space has the form of a tensor product, then the representation of the transformations of the solution space must have the form of a product representation? Because, we are looking at 6 data points here, and indeed, $\mathbb{R}^6 \cong \mathbb{R}^3 \otimes \mathbb{R}^2$, but surely Georgi cannot be suggesting that every system with 6 data points must transform as a product representation of a 3 dimensional representation and a 2 dimensional representation? Or is he?

Specifically, given a physical system, akin to the one in Georgi, or in the most general case, how can one tell whether the representation of the system must be a product representation or not, and how does one tell specifically what the "factor representations" must be?

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  • $\begingroup$ This seems more like a physics question, but let me take a stab. You haven't specified what acts on the $\mathbb{R}^2$ component. My guess is that this should be seen as an $GL_2$ representation. As such, we realize $\mathbb{R}^6=\mathbb{R}^3\otimes \mathbb{R}^2$ as a $S_3\times GL_2$ rep, making the tensor decomposition clear. $\endgroup$ – David Hill Nov 18 '16 at 0:50
  • $\begingroup$ My understanding of the text in Georgi has it that it is $S_3$ that acts upon the $\mathbb{R}^2$ product as well. Strictly speaking, Georgi, if I've understood him, has it that the $S_3$ group acts on the entire system, but he isn't clear as to why it would act on the different "factor spaces" distinctly and separately. Does this help? $\endgroup$ – StormyTeacup Nov 20 '16 at 16:43
  • $\begingroup$ This interpretation doesn't make much sense to me. There is an action of $S_3$ on $\mathbb{R}^2$ (as $S_3$ is the symmetry group of an equilateral triangle). It is unclear why this symmetry would leave the normal modes invariant That's a computation I would need to see. $\endgroup$ – David Hill Nov 21 '16 at 2:41
  • $\begingroup$ If I understood Georgi (and your question) correct, when you actually write out the rotations (in $GL_2$), you find these to be a two-dimensional representation of $S_3$. $\endgroup$ – StormyTeacup Nov 22 '16 at 8:00
  • $\begingroup$ But actually, perhaps you can help be by informing me why different groups would be acting on different components of the vector space? How can one tell that when that is the case in general? $\endgroup$ – StormyTeacup Nov 22 '16 at 8:01

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