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I am self studying quantum field theories and as many of you know this can not be properly done without at least some knowledge of Lie groups and the corresponding algebra.

I am reading the QFT text by Michele Maggiore “A modern introduction to quantum field theory”. In his chapter on Lie groups (I know, a single chapter is far from enough) he defines a representation of the Lie group as an operation that assigns a linear operator to each element within the Lie group.

My question is, does all Lie groups has such a (linear) representation? Or is Michele Maggiore restricting his discussion on Lie groups to that of matrix Lie groups without explicitly stating so?

This might be trivial question, but I am new to Lie groups and is just seeking some clarification on the topic.

William Crawford
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For any Lie group $G$ and any vector space $V$, there is a representation of $G$ which sends every element of $G$ to the identity operator on $V$. So, every Lie group has a representation.

In any case, though, I think you are kind of missing the point. We aren't interested in "groups which have a representation". Rather, we are interested in studying the many different representations which a single group may have. So even though a matrix Lie group "comes with" a canonical representation (since every element already is a matrix), we are still interested in other representations of such a group.

Eric Wofsey
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Presumably you are wondering whether every Lie group has a faithful finite-dimensional linear representation. Or in other words: is every Lie group realizable as a closed subgroup of $\mathrm{GL}(V)$ for some finite dimensional vector space $V$?

This is true for compact Lie groups (by the Peter-Weyl theorem) but false in general: for instance, the universal cover of $\mathrm{SL}_2(\mathbf{R})$ has no faithful finite-dimensional representation.

Stephen
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  • I am very curious why you said "is every Lie group realizable as a closed subgroup [...]." Does a faithful representation imply the closure property? – Maximal Ideal Feb 26 '20 at 23:48
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    @MaximalIdeal "Does a faithful representation imply the closure property?" No, certainly not. For instance consider the faithful representation of the one-dimensional Lie group $\mathbf{R}^1$ on $\mathbf{C}^2$ given by $\alpha \mapsto \mathrm{diag}(e^{2 \pi i a \alpha}, e^{2 \pi I \alpha})$ for an irrational real number $a$. This is of course a smooth faithful representation, but its image is dense in the compact torus $S^1 \times S^1$. – Stephen Feb 27 '20 at 11:39
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Your confusion is reasonable. First, as @DavidHill noted, we can always let a Lie group act by Adjoint on its Lie algebra, so there is at least one quite non-trivial linear representation. But, yes, this representation may map much of the Lie group to the trivial operator, if the center is large.

A question that may be underlying your question is about faithful linear representations, that is, where the group maps injectively/one-to-one to the linear transformations on the vector space. The question of whether a given abstract Lie group admits a faithful (finite-dimensional) linear representation is subtler, and the answer is not always "yes".

But compact Lie groups, for example, do always admit faithful linear representations, as was known already to Peter and Weyl and others almost 100 years ago.

And, yes, matrix Lie algebras behave somewhat more predictably that more general families.

paul garrett
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