Two notions of $SO(3)$'s irreps - what's the connection between them? While learning about their applications in physics and maths, I have gained three different notions of the irreps of the 3-dimensional rotation group (parametrized by spin $n\in\mathbb{N}$):


*

*(This notion turns out to be wrong) The most intuitive notion of irrep $n$ is $\exp(-in\vec{\omega}\cdot\vec{J})$, which operates on $\mathbb{R}^3$. For $n=0$ this is the identity, for $n=1$ it is the fundamental representation, and for higher $n$ it is like $n=1$, except that it "is rotating $n$ times faster than in the fundamental representation".

*The spherical harmonics of order $l$ are a basis of the irrep $n$ for $n=l$. This must have something to do with the fact that they are rotation symmetric for certain z-axis rotations with angle $\frac{2\pi}{n}$.

*Representations on 3D tensors of order $n$. For $n=0$ it is as always the trivial, one-dimensional irrep, for $n=1$ it is the fundamental one that acts on 3-vectors, and for $n=2$ it is the five-dimensional representation that acts on symmetric traceless 3x3 matrices.


I hope these notions are all correct, if not, please comment.
My question is, what's the connection between these three notions? For example, why is $\exp(-2i\vec{\omega}\cdot\vec{J})$ equivalent to the symmetric traceless 3x3 matrices? What's the exact connection between spherical harmonics and representations?

EDIT: Unfortunately, my understanding of these three notions is limited. I cannot describe the representations in detail (with vector space and operators). If I could do that, then most likely I would not ask this question. Unfortunately, many (physics) authors just say

XYZ is the $n$-th irrep of $SO(3)$

without giving any details about the representation, and the inexperienced reader does not know whether XYZ is the vector space or the operator of the representation.
So, I'm going to list a few sources, to make it clear how I got these notions. Hopefully someone is able to connect them.


*

*First bullet point: I don't know the exact source of this, but maybe I got this notion from the two-dimensional case, where the expression $\exp(-in\theta)$ is an irrep of $SO(2)$. Source: Robert Griffiths' Group Theory Essentials, page 10, formula (27). Maybe this does not extend to the 3D case.

*Second bullet point: Wikipedia.

*Third bullet point: For example, the last paragraph in this answer to a question of mine on PSE.


My question stays the same: What's the connection between these notions (if they are correct)?

EDIT 2: Turns out, bullet point 1 is definitely wrong.
Still looking for a connection between bullet points 2 and 3. Why are spherical harmonics a representation of $SO(3)$? How are the symmetric traceless 3x3 matrices an irrep of $SO(3)$?
 A: Based on some exchange of comments and some edits to the question, the question appears to be mainly directed at an explanation in what sense the spherical harmonics and the symmetric traceless $3\times3$ matrices form irreducible representations of $SO(3)$.
For the traceless $3\times3$ matrices:
Consider a matrix $A$ that represents a quadratic form $x^\top Ax$ for vectors $x\in\mathbb R^3$. If we rotate to $x'=Rx$ with some rotation matrix $R$, we must transform $A$ to $A'=RAR^\top$ so that $x'^\top A'x'=x^\top Ax$.
This is a linear transformation of matrices $A$, which form a $9$-dimensional vector space $\mathbb R^{3\times3}$, and it's a group action of the rotation group, since successively acting with two rotation matrices $R$ and $S$ yields $SRAR^\top S^\top=(SR)A(SR)^\top$, the action of the product $SR$. This action  defines a $9$-dimensional linear representation of $SO(3)$ on $\mathbb R^{3\times3}$. (This is a tensor product of two $3$-dimensional representations, but never mind if that doesn't mean anything to you.)
To find the character of this representation, consider
$$
A'_{il}=\sum_{jk}R_{ij}A_{jk}R_{lk}\;.
$$
The coefficient of $A_{il}$ in $A'_{il}$ is $R_{ii}R_{ll}$, so, denoting the linear map $A\to RAR^\top$ by $\rho(R)$, we have
\begin{align}
\def\tr{\operatorname{tr}}
\tr\rho(R)&=\sum_{il}R_{ii}R_{ll}\\
&=\left(\sum_iR_{ii}\right)^2\\
&=\left(\tr R\right)^2\\
&=(1+2\cos\theta)^2\\
&=1+4\cos\theta+4\cos^2\theta\\
&=3+4\cos\theta+2\cos2\theta\\
&=(1)+(1+2\cos\theta)+(1+2\cos\theta+2\cos2\theta)\;,
\end{align}
where $\theta$ is the rotation angle of $R$. (More generally, the character of a tensor product of two representations is the product of their characters.)
So, since the irreducible characters of $SO(3)$ are $1$, $1+2\cos\theta$, $1+2\cos\theta+2\cos2\theta$, ..., this tells us that our $9$-dimensional representation splits into one $1$-dimensional, one $3$-dimensional and one $5$-dimensional irreducible representation.
To identify these, first note that the multiples of the identity matrix transform among themselves, $RIR^\top=I$, so they form the $1$-dimensional irreducible subrepresentation. Next, note that the antisymmetric matrices also transform among themselves, $A=-A^\top\rightarrow RAR^\top=-(RAR^\top)^\top$, and they form a $3$-dimensional subspace, so this is the $3$-dimensional irreducible subrepresentation.
That leaves the orthogonal complement (orthogonal with respect to the Frobenius product) of those two spaces as the $5$-dimensional irreducible subrepresentation. The matrices orthogonal to all antisymmetric matrices are the symmetric matrices, and the matrices orthogonal to the identity are the traceless matrices, so this orthogonal complement is the $5$-dimensional space of traceless symmetric matrices. We can check that they transform among themselves: The trace is invariant, $\tr RAR^\top=\tr R^\top RA=\tr A$, and the symmetric matrices transform among themselves, $A=A^\top$ $\rightarrow$ $RAR^\top=(RAR^\top)^\top$.
For the spherical harmonics:
The spherical harmonics can be characterized as simultaneous eigenfunctions of the squared angular momentum operator $L^2$ and the $z$-component $L_z$ of the angular momentum operator. Since the components $L_\alpha$ commute with $L^2$, applying $L_\alpha$ to an eigenfunction of $L^2$ yields an eigenfunction of $L^2$ for the same eigenvalue: $L^2\psi=l(l+1)\psi$ $\rightarrow$ $L^2L_\alpha\psi=L_\alpha L^2\psi=L_\alpha l(l+1)\psi=l(l+1)L_\alpha\psi$.
Since the components of the angular momentum operator are the infinitesimal generators of rotations, this means that the eigenfunctions of $L^2$ with a given eigenvalue transform among themselves under rotations, so each eigenspace of $L^2$ affords a representation of the rotation group. It turns out that each of these representations is irreducible, and these irreducible representations are precisely all irreducible representations of the rotation group. The eigenspace of $L^2$ for given $l$ is the vector space spanned by the $2l+1$ spherical harmonics $Y_{lm}$ with $-l\le m\le l$.
The explicit forms of the transformations within each of these eigenspaces are complicated and not very enlightening; you can read about them e.g. here (Eq. ($63$) on p. $479$), here (Eq. ($6$) on p. $5622$) and here.
I'm sure I haven't answered half of what you wanted to know – I tried to guess what might be helpful to you – feel free to ask if questions remain.
