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I know that mathematicians define the fundamental representation of the algebra $\mathfrak{g}$ as the one whose highest weight is a fundamental weight (see e.g. a good answer here). If we follow this definition, then, naively, we should have at most $r = \operatorname{rank}\mathfrak{g}$ fundamental representations. Now, how do we further restrict this number? How to see the equivalence of some of these representations by looking at the weight lattice?

An example would be greatly appreciated. Say, $\mathfrak{su}(2)$ and $\mathfrak{su}(3)$ have same number of non-equivalent fundamental representations as their rank. What changes for $\mathfrak{su}(4)$? How to see that among representations built out of its three fundamental weights only two are non-equivalent?

In physics textbooks the algebras are defined through matrices. With such a definition it's implied that $\mathfrak{su}(n), n\geq 3$ algebras have only two fundamental representations - the defining one, and the one conjugated to it. Say, $4$ and $\overline{4}$ for $\mathfrak{su}(4)$. How does it coincide with the definition given in terms of the weight vectors?

Thanks and apologies for bad math language if used :)

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mavzolej
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    A complex semisimple lie algebra has as many fundamental weights as its rank. – Mariano Suárez-Álvarez Jan 23 '17 at 07:22
  • Notice that that is quite clearly stated in the answer you linked to. – Mariano Suárez-Álvarez Jan 23 '17 at 07:24
  • What would you say about the equation (C6) here? Why are there only two fundamental representations for $\mathfrak{su}(4)$? – mavzolej Jan 23 '17 at 07:26
  • It may well be that physicists have a different meaning for a fundamental representation. I am not sure that in algebra any special emphasis is given to the collection of irreducible representation whose highest weight happens to be a fundamental weight. Anyway, the number of fundamental weights is equal to the rank of the root system. It sounds like we are talking apples and oranges here. Also observe that $\mathfrak{su}(2)$ only has a single fundamental representation (all in accordance with the algebra). – Jyrki Lahtonen Jan 23 '17 at 07:33
  • In the notation from your link the representations $D^{100}$, $D^{010}$ and $D^{001}$ are the fundamental representations of $\mathfrak{su}(4)$ according to the definition that an irreducible representation is fundamental, iff its highest weight is. – Jyrki Lahtonen Jan 23 '17 at 07:36
  • My bad, for $n \geq 3$, of course, since for $\mathfrak{su} (2)$ we can occasionally relate the complex conjugated representations. – mavzolej Jan 23 '17 at 07:38
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    @mavzolej This is a clash of terminology. In physics literature "fundamental" often simply means what is also sometimes called the "defining" representation. So physics literature talks about the fundamental representation. Note that in the pdf you linked they didn't say two fundamental representations. They talked about only one fundamental representation and the other one was the conjugate representation (called the antifundamental). In math terminology number of fundamental representations = rank, as pointed out in the other comments. – nio Jan 23 '17 at 07:40
  • You should notice that the answerer to the linked question states: I'm assuming that the physics terminology is consistent with the mathematics terminology (which maybe is a big assumption). I don't think that the various people you refer to agree on which representations should be called fundamental. If they mean different things, then it is hardly a surprise that different people come up with different numbers of fundamental representations. – Jyrki Lahtonen Jan 23 '17 at 07:41
  • Oh my God, could you please clarify the case with $\mathfrak{su}(4)$? What are the dimensions of these $D^{100},~D^{010},~D^{001}$ representations? Can they all be obtained from one another by means of the Weyl transformations? – mavzolej Jan 23 '17 at 07:45
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    That physicist calls $D^{100}=4$, $D^{001}=\overline{4}$ and $D^{010}=8$. I don't know what a Weyl transformation is, but those representations are pairwise non-isomorphic (which is kind of the point). The two 4-dimensional ones are duals of each other, the 8-dimensional one is isomorphic to its own dual. I don't how physicists decide which one should be called $4$ and which is $\overline{4}$. In the case of the defining rep you can do that easily, but in general? – Jyrki Lahtonen Jan 23 '17 at 07:57
  • I meant some reflection from the Weyl group... Incredible - so, in the mathematical terminology there may be few fundamental representations which even have different dimensions.... – mavzolej Jan 23 '17 at 08:02
  • I guess, I should delete now my question in order to avoid the further confusion. – mavzolej Jan 23 '17 at 08:02
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    Actually I think there is an error in that formula C.6. The representation $D^{010}$ should be 6-dimensional. That is because it is the wedge product of (either) 4-dimensional representation with itself, and $\binom 4 2=6$. The dimension formula looks right, the author just plugged in wrong numbers. IIRC $SU(4)$ is not really used in elementary particle physics, so no wonder the author would not remember the dimensions of its irreducible reps by heart :-) – Jyrki Lahtonen Jan 23 '17 at 08:43
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    So if $x_1,x_2,x_3,x_4$ span the 4-dimensional representation (or carry it as physicists are wont of saying) of highest weight $\lambda_1$, then the wedge products $x_i\wedge x_j, 1\le i<j\le 4$ span that 6-dimensional representation with highest weight $\lambda_2$. The triple wedges span the rep with highest weight $\lambda_3$, but that is dual of the first one, IOW $\overline{4}$. – Jyrki Lahtonen Jan 23 '17 at 08:47
  • And, to reiterate, I am still not convinced that mathematicians would call these reps the fundamental representations. Possibly as a shorthand for representations with a highest weight that happens to be a fundamental weight, but that does not ring a bell. I guess that Eric Korman's answer was an attempt to try to make sense of that other question. – Jyrki Lahtonen Jan 23 '17 at 08:49
  • See Urs Schreiber's comment under Korman's post. I am inclined to think that he got it right. – Jyrki Lahtonen Jan 23 '17 at 08:51

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