A linear representation of a classical Lie group $G$ is defined by $\rho:g \to GL(V)$ where $g \in G$ is a group element and $V$ is the representation space. The dimension of the representation space(which is a linear vector space) is usually called the dimension of the representation.

The adjoint representation $Ad:G\to GL(\mathfrak{g})$ is defined via conjugation of an algebra element $A \in \mathfrak{g}$ i.e via $gAg^{-1}$ where $\mathfrak{g}$ is the Lie algebra. So the dimension of the representation is the dimension of the algebra which is $n^2-1$ for example for $SU(n)$.

Consider the defining/standard representation which is defined via a inclusion map to $GL(n,\mathbb{C})$. Then the dimension of the representation should be given by dimension of $\mathbb{C}^n \cong \mathbb{R}^{2n}$ which is $2n$. However we know that the fundamental representation and defining representation happens to be the same for classical Lie groups and fundamental representations are $n$ dimensional for $SU(n),SO(n)$ and $Sp(n)$. Why is this mismatch arising?

One might be inclined to claim that $2n$ is simply the real dimension and $n$ is the complex dimension (pointed out in a short discussion at the PSE chatroom---"The h bar"). But I fail to understand this because looking at, for example the $SU(2)$ decomposition $$2 \otimes 2 =3 \oplus 1$$ one is (probably?) talking about the complex dimension of the $2$ i.e. defining representation on LHS and real dimension of the ($3$) i.e. adjoint representation on the RHS! Surely, we need $4$ not $2$ real numbers to describe the elements(spinors) on which these representation act. Do the numbers in the $SU(2)$ decomposition has nothing to do with the dimension of the representation but simply the "size" of square matrices denoting the representations? By the way I don't view $(2s+1)$ as the definition of dimension. I view this merely as the relation of dimension to spin (just a very important parameter in physics) but not as a "definition" of dimension.

Summarizing, I want an answer which clarifies the usage of the word dimension from the perspective of dimension of some representation space, so that using that definition one can get dimension of both the defining and adjoint representations (of all i.e. both classical and "non-classical" Lie groups) correctly. Also, it is requested that the answerer also kindly points out the mistakes in the definitions I gave above, if possible.

P.S.--- Mathematics might be a better home for this question---but I wanted to post it in PSE because it might be the case that physicist's definitions and the mathematician's ones are not entirely the same in a topic like this.


1 Answer 1


The usual definition of dimension of a vector space explicitly mentions that one needs a field $F$ and a vector space $V$ over that field. The fact that the field of complex numbers $\mathbb{C}$ might be considered as a 2-dimensional vector space is also discussed:

... we have $\dim _{\mathbb {R} }(\mathbb {C} )=2$ and $\dim _{\mathbb {C} }(\mathbb {C} )=1.$ So the dimension depends on the base field.

Although I'd say that it is, indeed, true that physicists mostly never concern themselves with which base field they are talking about.

  • $\begingroup$ In the example $2 \otimes 2 =3 \oplus 1$ example, the fields are different on the two sides of the equality? $\endgroup$
    – Sanjana
    Oct 20, 2023 at 10:25

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