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The matrix elements for the $(2l+1)$-dimensional irreducible representation of SO(3) are given by:

$D_{m',m}^l(\phi_1,\Phi,\phi_2)=[i^{m'-m}\sqrt{(l+m')!(l-m')!(l+m)!(l-m)!}$ $\times(\frac{1+\cos\Phi}{2})^l (\frac{1-\cos\Phi}{1+\cos\Phi})^{(m'-m)/2}$ $\times\sum\limits_k \frac{(-1)^k}{(l+m-k)!(l-m'-k)!(m'-m+k)!k!}$ $\times(\frac{1-\cos\Phi}{1+\cos\Phi})^k] e^{im'\phi_2}e^{im\phi_1}$

My question is, how can there be multiple irreducible representations of SO(3) each of which has a different dimension? I thought that by definition an irreducible representation is one which has no non-trivial invariant subspace. I know that SO(3) can be represented by the group of 3x3 symmetric matrices of determinant = 1 (I thought this had something to do with the fact that it takes 3 parameters to describe a 3D rotation). If I have a larger matrix (e.g. 5x5) doesn't it have to have a non-trivial invariant subspace (e.g. of dimension 2), and therefore be reducible?

The application here is the expansion of functions using generalized spherical harmonics, which is accomplished (per the Peter-Weyl theorem) using an infinite linear combination of the matrix elements of the irreducible representationS (emphasis on the S) of SO(3).

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    Are you familiar with the representation theory of _finite_ groups? There are plenty of examples there. I don't really understand your comment; why must there be an invariant subspace of dimension $2$? (The irreducible representation of dimension $5$ is not the direct sum of the standard representation and two copies of the trivial representation; it is exactly what it says on the box, an irreducible representation of dimension $5$.)2011-05-19
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    Could you elaborate on why it seems to you that there must be such a non-trivial invariant subspace? It seems like you're basing this on some relationship between the $3\times3$ matrices and the $5\times5$ matrices, e.g. that the former have to be contained in the later or something like that. That's not the case; the $5$-dimensional representation is a whole new set of matrices, with no immediate connection to the $3$-dimensional representation.2011-05-19
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    @Qiaochu Yuan: I guess I am erroneously making some sort of connection between irreducible representations and minimal spanning sets (i.e. bases). Is there no such connection?2011-05-19
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    @okj: That's a very general question. A representation is a representation on some vector space, that vector space has a basis, and if you express the linear transformations of the representation in that basis, you get a corresponding matrix representation -- so very generally speaking there is a connection, but you'd have to elaborate what sort of connection you're presuming that would lead to your ideas about invariant subspaces.2011-05-19
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    @okj: I don't understand what bases have to do with this.2011-05-19
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    @joriki: I guess I'm confusing the basis of the vector space ($R^3$), with a basis for the representation (a $(2n+1)\times (2n+1)$ matrix). SO(3) are 3D rotations, so naturally I would assume that representatives of SO(3) would act on vectors of length 3, with usual matrix/vector multiplication. So what I don't understand is, how can I have a 5x5 (or any other size besides 3x3) matrix representation of SO(3) that is non-singular, and how would it multiply a vector that is 3x1?2011-05-19
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    @okj: You need to abstract from the $3$-dimensional representation (also called the "defining representation"). A representation is quite generally every map that assigns to each group element some linear transformation (or matrix, if you prefer) such that the mapping commutes with composition. These can be linear transformations on arbitrary vector spaces. It's probably easier to understand for the $1$-dimensional trivial representation, which assigns $1$ to every group element. Do you understand how this operates on $1$-dimensional vectors?2011-05-19
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    @joriki: thanks for the color warning. I think okj has expressed his concern clearly now. Maybe someone can describe how the red, green, blue correspond to the x,y,z of QY's reply, and then mention how SO3 also must act on x*x and y*y, oh my, and so give the next irrep.2011-05-19
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    @joriki: so if I understand what you're saying, each distinct representation of a group can act on a different vector space? so that for a single group (SO(3) in my case) the 3x3 representation will act on a 3-dimensional vector space, while the 5x5 representation would act on a 5 dimensional vector space?2011-05-19
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    @Jack Schmidt: you posted an answer that was quite helpful, but it disappeared... where did it go?2011-05-19
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    @okj: Yes, that's exactly right. Not only *can* it act on a different vector space, it must, since they have different dimensions.2011-05-19
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    @joriki: so is that to say then that our common interpretation of SO(3) as 3D rotations is not abstractly true? I would assume that any 3D rotation acting on $R^5$ would have to leave the basis vectors of two dimensions unchanged (just as a 2D rotation in $R^3$ always leaves one dimension unchanged--which is an eigenvector corresponding to an eigenvalue of +1).2011-05-19
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    @okj: undeleted, incorporating a fix from joriki, and hopefully explaining the polynomial thing Qiaochu Yuan mentioned in his reply.2011-05-19
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    @okj: When we say $SO(3)$ "acts" on a vector space, that's not meant in the sense that 3D rotations act on that vector space as rotations. It means that we map each element of $SO(3)$ to some linear transformation of the vector space, such that the map respects the group multiplication, and it's that linear transformation that "acts" on the vector space -- saying that $SO(3)$ "acts" on the space is just a shorthand for that. So it's perfectly OK to regard $SO(3)$ as the group of all 3D rotations; the problem only begins when you imagine these acting directly as rotations on other spaces.2011-05-19
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    @joriki: Thank you, so what I'm understanding from your comment as well as Jack's is that SO(3) always acts on configuration space (i.e. $R^3$) with the representation of 3x3 matrices. However, we can form function spaces of arbitrary degree, and SO(3) can act directly on these function spaces (e.g. $R^6$ with basis functions xx,xy,yy,xz,yz,zz), but must do so using a representation of equivalent dimension (and there are irreducible representations for any dimension), though their effect on function space, e.g. $R^6$, will not be a 3D rotation in the sense that it is for configuration space.2011-05-19
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    @okj: Yes, that's correct. However, I wouldn't make such a strong distinction between configuration spaces and function spaces. For instance, in physics a $2$-dimensional representation of $SO(3)$ acts on spinors -- would this be a function space or a configuration space? Also, the function space spanned by $x$, $y$ and $z$ transforms under the same representation as the ordinary $3$-dimensional configuration space.2011-05-19

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$\mathrm{SO}(3)$ is a wonderful group, but it quickly got bored rotating vectors in space. It wanted to be used to help solve differential equations, and so desperately wanted to act on functions. Then one day, it realized it could act on functions by acting on their input.

For instance, it contains a 90 degree clockwise rotation in the $x$-$y$ plane. It takes $(x,y,z)$ and replaces it with $(y,-x,z)$.

Well, the function $2x+3y+7z$ can also be rotated! We just replace $x$ by $y$, $y$ by $-x$, and leave $z$ alone: $2x+3y+7z$ is rotated into $2y+3(-x)+7z = -3x + 2y + 7z$. If we consider the action on all these linear $Ax+By+Cz$ it becomes clear that $\mathrm{SO}(3)$ acts by 3×3 matrices on the vector space with basis $\{x,y,z\}$.

Yay, $\mathrm{SO}(3)$ can act on functions now, and it acts on those linear functions as 3×3 matrices.

What about quadratics? Well we could have $2x^2 + 3xy + 7y^2$. That same 90 degree rotation takes it to $2(y)(y) + 3(y)(-x) + 7(-x)(-x) = 7x^2 - 3xy + 2y^2$. Yay, now $\mathrm{SO}(3)$ acts as 6×6 matrices on the quadratic polynomials $Ax^2+Bxy+Cy^2+Dxz+Eyz+Fz^2$ spanned by $\{ x^2, xy, y^2, xz, yz, z^2 \}$.

It turns out though that $\mathrm{SO}(3)$ doesn't swirl these functions around very thoroughly. If you take a function, and rotate its input, all it does is rotate the laplacian. In particular, notice that it takes any harmonic polynomial (one with 0 laplacian) to a harmonic polynomial. For quadratic polynomials, this is particularly easy to describe:

When you use $\mathrm{SO}(3)$ to rotate $Ax^2+Bxy+Cy^2+Dxz+Eyz+Fz^2$, it leaves the laplacian, $A+C+F$, alone. So it will never rotate $x^2$ into $xy$. In fact, it leaves the 5-dimensional space spanned by $\{ x^2-y^2, y^2-z^2, xy, xz, yz \}$ invariant. If we write things in these coordinates, then we get $\mathrm{SO}(3)$ acting as 5×5 matrices. In fact it acts irreducibly.

What happened to the 6th dimension? Well, it is spanned by a very silly function, the sphere: $x^2+y^2+z^2$. If you rotate the sphere, you get the sphere. On this one dimensional space, $\mathrm{SO}(3)$ is represented by 1×1 matrices. Well, actually, matrix. Every single rotation in $\mathrm{SO}(3)$ acts as the identity matrix $[1]$, since every rotation leaves the sphere alone.

If we use colors, like pink, lime, and periwinkle to describe vectors, then a polynomial like $5x^2-5y^2 + 6x + 3$ is composed of a pink term, $5x^2-5y^2$, where $\mathrm{SO}(3)$ acts in a 5×5 manner, a lime term, $6x$, where $\mathrm{SO}(3)$ acts in a 3×3 manner, and a periwinkle term, $3$, where $\mathrm{SO}(3)$ acts in a 1×1 manner.

A polynomial like $7x^2+5y^2$ is a little trickier to see its colors (it is a good thing $\mathrm{SO}(3)$ is so clever), $7x^2+5y^2 = 4(x^2+y^2+z^2) + 3(x^2-y^2) + 4(y^2-z^2)$. The fist term $4(x^2+y^2+z^2)$ is periwinkle where $\mathrm{SO}(3)$ acts in a 1×1 manner, but the next two terms $3(x^2-y^2) + 4*(y^2-z^2)$ are pink where $\mathrm{SO}(3)$ acts in a 5×5 manner.

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    Since the question (especially the matrix elements) looks like it might be motivated by physics, it may be worthwhile to point out that your colour metaphor is different from how colour is used in connection with quantum chromodynamics, where incidentally the symmetry group is also $SO(3)$, and the colours red, green and blue are used to label the three basis vectors *within* the $3$-dimensional irreducible representation.2011-05-19
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    Does anyone know how to do color on StackExchange answers? A google search brought up a ton of questions about how to do color in every other medium known to man.2011-05-19
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    @Jack since the markup is a subset of latex, the command syntax- `$\color{color-you-want}{text}$` will introduce color in the math expression (i.e. within dollar signs), if you want colored text - place the text within a math-mode `\text{text-goes-here}` command2011-05-19
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    @Jack: You can do it inside formulas like this: `$\color{red}{r}\color{green}{g}\color{blue}{b}$`: $\color{red}{r}\color{green}{g}\color{blue}{b}$. I don't know whether it's possible outside formulas.2011-05-19
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    @Jack Schmidt: Thank you very much! I think this answer is very helpful. As I stated, I'm interested in spherical harmonics, so if I'm interpreting your response correctly it seems like each degree of harmonics is like a new order of polynomial (your example of x,y,z vs. xx,xy,yy,xz,yz,zz) requiring a representation of SO(3) of a matching dimension to the number of basis functions of that degree. E.g. $Y_0^m$ has only one basis function so would require a 1x1 representation of SO(3), whereas $Y_2^m$ has 5 basis functions, requiring a 5x5 representation of SO(3). Is this interpretation correct?2011-05-19
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    +1 for periwinkle! (and generally the enjoyable style of this answer.) Too bad `$\color{periwinkle}{3}$` doesn't work...2011-05-19
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    @okj: Yes, that's correct. Note, however, that in your first example, the six second-order polynomials correspond to a *reducible* representation (as Jack explained, $1 + 5$), whereas the $Y_l^m$ are specifically designed to form *irreducible* representations of $SO(3)$ for each $l$. By the way, perhaps not surprisingly, the $1$- and $5$-dimensional representations that the representation of the six second-order polynomials reduces to are precisely the ones corresponding to the spherical harmonics for $l=0$ and $l=2$.2011-05-19
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    Hey Jack, don't anthropomorphize SO(3): it hates it!2011-05-20
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    @joriki I thought the symmetry group in QCD was SU(3) emphasis on the U?2018-11-15
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That's the most confusing way to write down the irreducible representations of $\text{SO}(3)$ I've ever seen. Here's a much less confusing one: the irreducible representation of $\text{SO}(3)$ of dimension $2n+1$ is precisely its action on the space of harmonic polynomials in three variables $x, y, z$ of degree $n$ (given by extending its action on $x, y, z$ multiplicatively). Maybe this will help you play with these representations and see why they don't behave the way you think they do.

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Dear okj, in the $SO(3)$ case, and perhaps at least morally even more generally, you may imagine the higher-dimensional irreducible representations as tensors. In particular, the 5-dimensional representation is a symmetric traceless tensor $T$. It satisfies $$T_{ij} = T_{ji}, \quad \sum_{i=1}^3 T_{ii} = 0 $$ The first condition reduces 9 elements of the matrix to 6 independent ones; the vanishing of the trace reduces 6 to 5. Now, if an $SO(3)$ matrix $M$ acts on vectors via $$ \vec v \to M \vec v ,$$ then it acts on $T$ via $$ T \to M T M^{T}.$$ Note that it preserves the symmetry of $T$ as well as its vanishing trace. However, it's equally clear that there is no linear combination of the five entries in $T$ that would be invariant under all $SO(3)$ transformations. So there is no singlet, and because $3+1+1$ and $1+1+1+1+1$ are the only possible ways how to decompose a 5-dimensional representation to representations you know, it follows that none of them works and there is no triplet (vector), either.

More general representation of $SO(3)$ may also be written as tensors with some constraints. The constraints are needed to make them irreducible.

For more general groups, both finite and Lie groups, the more complicated representations are not that easily written as tensors, but in principle, it's always possible to find any irrep in the tensor product of "fundamental representations".

Here is an important point that may be confusing you. One may build more complicated representations of a group as $$ V_1\oplus V_2$$ which is manifestly reducible but one may also build them as $$ V_1\otimes V_2, $$ the tensor product. This is like creating object with many indices. There is no simple link between the direct sum and the tensor product. The tensor product typically contains completely new, higher-dimensional ireducible representations than $V_1$ and $V_2$.