There is a conclusion on my textbook:

All semisimple algebraic groups of type $F_4$ are isomorphic.

I was confused because of the fundamental groups.

Is it true that two semisimple algebraic groups could be isomorphic only when their fundamental groups are isomorphic? So, all semisimple algebraic groups of type $F_4$ have isomorphic fundamental groups?

But if $\Lambda$ denotes the weight lattice, and $\Lambda_r$ denotes the root lattice, then under calculation, I found that $[\Lambda:\Lambda_r]=29$, so the fundamental group may have order $1$ or $29$... Am I wrong?

Moreover, is there any connection between the fundamental group (of the semisimple algebraic group) and the diagram automorphism of the Dynkin diagram of the corresponding root system? The semisimple algebraic groups of type $F_4$ have only one fundamental group (up to isomorphism), because the Dynkin diagram of $F_4$ has no nontrivial diagram automorphism? Is this true? And if it is, why?

Thank you very much.

[I think this has a deep connection to Lie algebras, so I will give two tags to this question.]

  • $\begingroup$ Answer to myself: the number of diagram isomorphisms and $[\Lambda: \lambda_r]$ don't need to coincide. In case of $A_1$, the first number is $1$, and second $2$. So, calculation is needed!! $\endgroup$ Sep 10, 2011 at 12:43

1 Answer 1


I'm afraid something has gone wrong in your calculation. A table of fundamental weights on page 69 of Humphreys' book "Introduction to Lie Algebras and Representation Theory", Springer GTM series #9 (my copy printed in 1980, so page number may vary a bit) lists the fundamental weights as integral linear combinations of roots. Thus the weight lattice is a subset of the root lattice. Hence the root and weight lattices of the simple Lie algebras of this type coincide (the other inclusion being always true). This shows that in this case the fundamental group is, indeed, always trivial.

Section 13.1 in loc.cit. also lists the index of the root lattice as a subgroup of the weight lattice as 1. It should be relatively straightforward to verify this from any geometric description of the $F_4$ root lattice as well. Please check your calculations. If you cannot spot the mistake, post it here, and we can work on it. It may be just an error in computing the value of a 4x4 determinant.


Using the identification of the simple roots with vectors of $\mathbf{R}^4$ from this page I got $\lambda_1=(1,0,0,-1)=2\alpha_1+3\alpha_2+4\alpha_2+2\alpha_1$, $\lambda_2=(1,1,0,-2)=3\alpha_1+6\alpha_2+8\alpha_3+4\alpha_4$, $\lambda_3=\frac12(1,1,1,-3)=2\alpha_1+4\alpha_2+6\alpha_3+3\alpha_4$ and $\lambda_4=(0,0,0,-1)=\alpha_2+2\alpha_2+3\alpha_3+2\alpha_4$. Here $\alpha_1,\alpha_2$ are the long basic roots and $\alpha_3,\alpha_4$ are short. Furthermore, $\lambda_1$ is also the highest long root, and $\lambda_4$ is the highest short root. Anyway, the fundamental weights clearly belong to the root lattice settling the question.

  • $\begingroup$ Thank you very much for your answer. Now I know where I was wrong. I didn't realize that that the value of the basic roots could be set according to the Cartan matrix, to make the calculation easier. Many thanks for your kind help~ $\endgroup$ Sep 10, 2011 at 12:37

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