This will perhaps be an unenlightening question, but here I go. Hopefully someone can varify my thoughts.


Considering Lie Group Theory and Representation Theory, for the case of the $SO(3)$, the special orthogonal group of $3\!\times\!3$-matrices, is the adjoint representation the same as the fundamental representation?


I think the should be, since the fundamental representation is defined to be

$$ \mathcal{D}(g) = g $$

which for $ O \in SO(3) $ is a $3\!\times\!3$-matrix, and hence acts on a vector space $V \subset \mathbb{R}^3$.

Moreover, the adjoint representation is defined to be the representation which acts on a vector space whoes dimension is equal to that of the group. That is, $U \subset \mathbb{R}^n$ where

$$ \dim(G) = n $$

We know that for the special orthogonal group

$$ \dim[SO(n)] =\frac{n(n-1)}{2} $$

So in the case of $SO(3)$ this is

$$ \dim[SO(3)] =\frac{3(3-1)}{2} = 3 $$

Thus we need the adjoint representation to act on some vectors in some vector space $W \subset \mathbb{R}^3$. That obvious choice to me is the $SO(3)$ matrices themselves, but I can't seem to find this written anywhere.


So to conclude, am I correct in thinking that the fundamental representation and the adjoint representation of $SO(3)$ are the same?

If so, are there any 'special' consequences to them being the same? It is generally the case that the fundamental representation and the adjoint representation of a group are not the same, to the best of my knowledge. Certainly they are defined as different things.


By the way, I am a maths-physics student, and do not have any pure-maths course on representation theory, only a 'for-physicists' one! This question has arisen in that vain.


Thank you!

  • 1
    $\begingroup$ Usually the fundamental representation and the adjoint representation have different dimensions. $\endgroup$ May 14, 2014 at 20:56
  • $\begingroup$ @QiaochuYuan : Yes, exactly. I guess that's what I'm asking about. Usually they act on different dimension vector spaces, and are quite clearly (to me) 'different' representations. However in this case there is the 'coincidence' they they have the same dimension, so I am wondering if that is all it is, a coincidence, or if indeed they are the same thing in this special case? Thanks. $\endgroup$
    – Flint72
    May 14, 2014 at 21:00

2 Answers 2


Yes. Actually the representation theory of $\text{SO}(3)$ is quite constricted: there is precisely one irreducible representation in each odd dimension $1, 3, 5, ...$ and that's it. In particular, there are only two $3$-dimensional representations, one of which is the fundamental / adjoint representation and the other of which is trivial.

I don't know if there are any special consequences of this. For $\text{SO}(n)$ for general $n$ the adjoint representation is the exterior square of the fundamental representation (exercise).


What you observed here (for SO(3), adjoint rep. = defining rep.) is indeed a very fine and remarkable property of SO(3). It explains, for example, the vector cross product in Lie-algebraic terms: the cross product R^3x R^3 --> R^3 is precisely the commutator of the Lie algebra, [,]: so(3)x so(3) --> so(3), i.e. the differential of the adjoint rep. of its Lie group! And it only works because vectors in R^3 can be identified with elements of the Lie algebra so(3) in a unique and SO(3)-equivariant way.

  • $\begingroup$ Sorry to quibble, but: "a unique $\operatorname{SO}(3)$-equivariant way" (that is, it's only unique when we postulate equivariance in advance); and, even then, "an $\operatorname{SO}(3)$-equivariant way, unique up to non-$0$ scalar multiples." $\endgroup$
    – LSpice
    Jul 2, 2023 at 4:00

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