Representations of the Special Orthogonal Group in Three Dimensions. This will perhaps be an unenlightening question, but here I go. Hopefully someone can varify my thoughts.
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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?
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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.
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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.
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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.
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Thank you!
 A: 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). 
A: 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.
