3.15. Lecture 15

3.15.1. Decomposing Sym2โก(โ„‚2)โŠ—Sym2โก(โ„‚2)\operatorname{Sym}^{2}(\mathbb{C}^{2})\otimes\operatorname{Sym}^{2}(\mathbb{C}% ^{2})

Example 3.15.1.

Let โ„‚2\mathbb{C}^{2} be the standard representation of SL2โก(โ„‚)\operatorname{SL}_{2}(\mathbb{C}) with weight basis ๐ž1{\bf e}_{1}, ๐žโˆ’1{\bf e}_{-1}. Consider V=Sym2โก(โ„‚2)โŠ—Sym2โก(โ„‚2)V=\operatorname{Sym}^{2}(\mathbb{C}^{2})\otimes\operatorname{Sym}^{2}(\mathbb{% C}^{2}). Let ๐ฏ2=๐ž12{\bf v}_{2}={\bf e}_{1}^{2}, ๐ฏ0=๐ž1โข๐žโˆ’1{\bf v}_{0}={\bf e}_{1}{\bf e}_{-1} and ๐ฏโˆ’2=๐žโˆ’12{\bf v}_{-2}={\bf e}_{-1}^{2} be weight vectors in Sym2โก(โ„‚2)\operatorname{Sym}^{2}(\mathbb{C}^{2}) corresponding to the weights 22, 0 and โˆ’2-2. Then the weights of VV are

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    44, multiplicity one, weight vector: ๐ฏ2โŠ—๐ฏ2{\bf v}_{2}\otimes{\bf v}_{2}.

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    22, multiplicity two, weight space: โŸจ๐ฏ2โŠ—๐ฏ0,๐ฏ0โŠ—๐ฏ2โŸฉ\left\langle{\bf v}_{2}\otimes{\bf v}_{0},{\bf v}_{0}\otimes{\bf v}_{2}\right\rangle.

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    0, multiplicity three, weight space: โŸจ๐ฏ2โŠ—๐ฏโˆ’2,๐ฏ0โŠ—๐ฏ0,๐ฏโˆ’2โŠ—๐ฏ2โŸฉ\left\langle{\bf v}_{2}\otimes{\bf v}_{-2},{\bf v}_{0}\otimes{\bf v}_{0},{\bf v% }_{-2}\otimes{\bf v}_{2}\right\rangle.

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    โˆ’2-2, multiplicity two, weight space: โŸจ๐ฏ0โŠ—๐ฏโˆ’2,๐ฏโˆ’2โŠ—๐ฏ0โŸฉ\left\langle{\bf v}_{0}\otimes{\bf v}_{-2},{\bf v}_{-2}\otimes{\bf v}_{0}\right\rangle.

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    โˆ’4-4, multiplicity one, weight vector: โŸจ๐ฏโˆ’2โŠ—๐ฏโˆ’2โŸฉ\left\langle{\bf v}_{-2}\otimes{\bf v}_{-2}\right\rangle.

Refer to caption
Figure 3.1. Decomposing Sym2โก(โ„‚2)โŠ—Sym2โก(โ„‚2)\operatorname{Sym}^{2}(\mathbb{C}^{2})\otimes\operatorname{Sym}^{2}(\mathbb{C}% ^{2}).

The weights of Sym2โก(โ„‚2)\operatorname{Sym}^{2}(\mathbb{C}^{2}) are {โˆ’2,0,2}\{-2,0,2\} and so the weights of Sym2โก(โ„‚2)โŠ—Sym2โก(โ„‚2)\operatorname{Sym}^{2}(\mathbb{C}^{2})\otimes\operatorname{Sym}^{2}(\mathbb{C}% ^{2}) are

{โˆ’2,0,2}+{โˆ’2,0,2}={โˆ’4,โˆ’2,โˆ’2,0,0,0,2,2,4}.\{-2,0,2\}+\{-2,0,2\}=\{-4,-2,-2,0,0,0,2,2,4\}.

This is the same as the set of weights of

Sym4โก(โ„‚2)โŠ•Sym2โก(โ„‚2)โŠ•โ„‚\operatorname{Sym}^{4}(\mathbb{C}^{2})\oplus\operatorname{Sym}^{2}(\mathbb{C}^% {2})\oplus\mathbb{C}

and so this is the required decomposition into irreducibles.

We can go further, and decompose VV into irreducible *sub*representations. This means finding irreducible subrepresentations of VV such that VV is their direct sum (not just isomorphic to their sum).

The copy of Sym4โก(โ„‚2)\operatorname{Sym}^{4}(\mathbb{C}^{2}) in VV has highest weight vector ๐ฏ2โŠ—๐ฏ2{\bf v}_{2}\otimes{\bf v}_{2}. We can find a basis by repeatedly acting with YY on it (writing โˆ\propto for โ€˜equal up to a non-zero scalarโ€™):

Yโข(๐ฏ2โŠ—๐ฏ2)\displaystyle Y({\bf v}_{2}\otimes{\bf v}_{2}) =2โข(๐ฏ2โŠ—๐ฏ0+๐ฏ0โŠ—๐ฏ2)โˆ๐ฏ2โŠ—๐ฏ0+๐ฏ0โŠ—๐ฏ2\displaystyle=2({\bf v}_{2}\otimes{\bf v}_{0}+{\bf v}_{0}\otimes{\bf v}_{2})% \propto{\bf v}_{2}\otimes{\bf v}_{0}+{\bf v}_{0}\otimes{\bf v}_{2}
Y2โข(๐ฏ2โŠ—๐ฏ2)\displaystyle Y^{2}({\bf v}_{2}\otimes{\bf v}_{2}) โˆ๐ฏ2โŠ—๐ฏโˆ’2+4โข๐ฏ0โŠ—๐ฏ0+๐ฏโˆ’2โŠ—๐ฏ2\displaystyle\propto{\bf v}_{2}\otimes{\bf v}_{-2}+4{\bf v}_{0}\otimes{\bf v}_% {0}+{\bf v}_{-2}\otimes{\bf v}_{2}
Y3โข(๐ฏ2โŠ—๐ฏ2)\displaystyle Y^{3}({\bf v}_{2}\otimes{\bf v}_{2}) โˆ6โข(๐ฏ0โŠ—๐ฏโˆ’2+๐ฏโˆ’2โŠ—๐ฏ0)โˆ๐ฏ0โŠ—๐ฏโˆ’2โŠ—๐ฏโˆ’2โŠ—๐ฏ0\displaystyle\propto 6({\bf v}_{0}\otimes{\bf v}_{-2}+{\bf v}_{-2}\otimes{\bf v% }_{0})\propto{\bf v}_{0}\otimes{\bf v}_{-2}\otimes{\bf v}_{-2}\otimes{\bf v}_{0}
Y4โข(๐ฏ2โŠ—๐ฏ2)\displaystyle Y^{4}({\bf v}_{2}\otimes{\bf v}_{2}) โˆ๐ฏโˆ’2โŠ—๐ฏโˆ’2.\displaystyle\propto{\bf v}_{-2}\otimes{\bf v}_{-2}.

These vectors are a basis for the copy of Sym4โก(โ„‚2)\operatorname{Sym}^{4}(\mathbb{C}^{2}) in VV.

Next, we find the copy of Sym2โก(โ„‚2)\operatorname{Sym}^{2}(\mathbb{C}^{2}) in VV. We start by looking for a highest weight vector of weight 2:

๐ฏ2โŠ—๐ฏ0โˆ’๐ฏ0โŠ—๐ฏ2{\bf v}_{2}\otimes{\bf v}_{0}-{\bf v}_{0}\otimes{\bf v}_{2}

does the trick. Hitting this with YY gives ๐ฏ2โŠ—๐ฏโˆ’2โˆ’๐ฏโˆ’2โŠ—๐ฏ2{\bf v}_{2}\otimes{\bf v}_{-2}-{\bf v}_{-2}\otimes{\bf v}_{2}, and doing so again gives ๐ฏโˆ’2โŠ—๐ฏ0โˆ’๐ฏ0โŠ—๐ฏโˆ’2{\bf v}_{-2}\otimes{\bf v}_{0}-{\bf v}_{0}\otimes{\bf v}_{-2} (up to scalar). These vectors are a basis for the copy of Sym2โก(โ„‚2)\operatorname{Sym}^{2}(\mathbb{C}^{2}) in VV.

Finally, we find the trivial representation โ„‚\mathbb{C} in VV. We need only find a weight vector of weight 0 which is killed by XX, and

๐ฏ2โŠ—๐ฏโˆ’2โˆ’2โข๐ฏ0โŠ—๐ฏ0+๐ฏโˆ’2โŠ—๐ฏ2{\bf v}_{2}\otimes{\bf v}_{-2}-2{\bf v}_{0}\otimes{\bf v}_{0}+{\bf v}_{-2}% \otimes{\bf v}_{2}

does the job. This vector spans a copy of the trivial representation.

3.15.2. Classification of irreducible SโขOโข(3)SO(3) representations

We have already seen that ๐”ฐโข๐”ฒ2\mathfrak{su}_{2} and ๐”ฐโข๐”ฌ3\mathfrak{so}_{3} are isomorphic (Problemย 1.5.2). We therefore have:

Theorem 3.15.2.

There is an irreducible two-dimensional representation VV of ๐”ฐโข๐”ฌ3\mathfrak{so}_{3} such that the irreducible complex representations of ๐”ฐโข๐”ฌ3\mathfrak{so}_{3} are exactly Symnโก(V)\operatorname{Sym}^{n}(V) for nโ‰ฅ0n\geq 0.

Proof.

By Theoremย 3.14.6, all irreducible representations of ๐”ฐโข๐”ฒ2\mathfrak{su}_{2} are of the form Symnโก(โ„‚2)\operatorname{Sym}^{n}(\mathbb{C}^{2}) for nโ‰ฅ0n\geq 0. By the isomorphism from Problemย 1.5.2, ๐”ฐโข๐”ฌ3โ†’โˆผ๐”ฐโข๐”ฒ2โІ๐”คโข๐”ฉ2,โ„‚\mathfrak{so}_{3}\xrightarrow{\sim}\mathfrak{su}_{2}\subseteq\mathfrak{gl}_{2,% \mathbb{C}}, we have that all irreducible representations of ๐”ฐโข๐”ฌ3\mathfrak{so}_{3} are isomorphic to something of this form. โˆŽ

We want to know which of these representations exponentiate to an irreducible representation of SOโข(3)\mathrm{SO}(3). For this, we revisit the connection with SUโข(2)\mathrm{SU}(2).

Let โŸจโ‹…,โ‹…โŸฉ\left\langle\cdot,\cdot\right\rangle be the bilinear form on ๐”ฐโข๐”ฒ2\mathfrak{su}_{2} defined by

โŸจX,YโŸฉ=โˆ’trโก(XโขY).\left\langle X,Y\right\rangle=-\operatorname{tr}(XY).
Lemma 3.15.3.

The form โŸจโ‹…,โ‹…โŸฉ\left\langle\cdot,\cdot\right\rangle is a positive definite bilinear form preserved by the adjoint action of SUโข(2)\mathrm{SU}(2).

Proof.

This is Problemย 1.7.3 where we note that AdgโกX=gโขXโขgโˆ’1\operatorname{Ad}_{g}X=gXg^{-1}. โˆŽ

Corollary 3.15.4.

There is an isomorphism SUโข(2)/{ยฑId}โ‰…SOโข(3).\mathrm{SU}(2)/\{\pm\operatorname{Id}\}\cong\mathrm{SO}(3).

Proof.

Choosing an orthonormal basis for ๐”ฐโข๐”ฒ2\mathfrak{su}_{2} with respect to โŸจโ‹…,โ‹…โŸฉ\left\langle\cdot,\cdot\right\rangle identifies the set of linear maps ๐”ฐโข๐”ฒ2โ†’๐”ฐโข๐”ฒ2\mathfrak{su}_{2}\to\mathfrak{su}_{2} preserving the inner product with SOโข(3)\mathrm{SO}(3). We therefore have a homomorphism

SUโข(2)โŸถSOโข(3)\mathrm{SU}(2)\longrightarrow\mathrm{SO}(3)

whose kernel is {gโˆˆSUโข(2)|Adg=Id}={ยฑId}\{g\in\mathrm{SU}(2)\,|\,\operatorname{Ad}_{g}=\operatorname{Id}\}=\{\pm% \operatorname{Id}\}. The derivative of this homomorphism is injective, otherwise there would be a one-parameter subgroup in the kernel of the group homomorphism, and since the Lie algebras have the same dimension we get an isomorphism of Lie algebras. The group homomorphism is therefore surjective, since exponentials of elements of ๐”ฐโข๐”ฌ3\mathfrak{so}_{3} generate the group SOโข(3)\mathrm{SO}(3). โˆŽ

Remark 3.15.5.

Let โ„,๐’ฅ,๐’ฆ\mathcal{I},\mathcal{J},\mathcal{K} be the elements of ๐”ฐโข๐”ฒ2\mathfrak{su}_{2} given by

12โข(01โˆ’10),12โข(0ii0),12โข(i00โˆ’i)\frac{1}{2}\begin{pmatrix}0&1\\ -1&0\end{pmatrix},\ \frac{1}{2}\begin{pmatrix}0&i\\ i&0\end{pmatrix},\ \frac{1}{2}\begin{pmatrix}i&0\\ 0&-i\end{pmatrix}

respectively. Up to scalar, these are an orthonormal basis for ๐”ฐโข๐”ฒ2\mathfrak{su}_{2}. Since the derivative of the above homomorphism SUโข(2)โ†’SOโข(3)\mathrm{SU}(2)\to\mathrm{SO}(3) is the adjoint map, computed with respect to this basis, we see that the induced isomorphism ๐”ฐโข๐”ฒ2โ†’๐”ฐโข๐”ฌ3\mathfrak{su}_{2}\to\mathfrak{so}_{3} takes:

โ„โŸผJx\displaystyle\mathcal{I}\longmapsto J_{x} =(00000โˆ’1010)\displaystyle=\begin{pmatrix}0&0&0\\ 0&0&-1\\ 0&1&0\end{pmatrix}
๐’ฅโŸผJy\displaystyle\mathcal{J}\longmapsto J_{y} =(001000โˆ’100)\displaystyle=\begin{pmatrix}0&0&1\\ 0&0&0\\ -1&0&0\end{pmatrix}
๐’ฆโŸผJz\displaystyle\mathcal{K}\longmapsto J_{z} =(0โˆ’10100000).\displaystyle=\begin{pmatrix}0&-1&0\\ 1&0&0\\ 0&0&0\end{pmatrix}.

It will be useful to know where this isomorphism takes the elements H,X,YH,X,Y of ๐”ฐโข๐”ฉ2,โ„‚=๐”ฐโข๐”ฒ2,โ„‚\mathfrak{sl}_{2,\mathbb{C}}=\mathfrak{su}_{2,\mathbb{C}}. For example, as H=โˆ’2โขiโข๐’ฆH=-2i\mathcal{K}, we see that it goes to โˆ’2โขiโขJz-2iJ_{z}. Or, for the lowering operator YY, we have

Y=(0010)=โˆ’(โ„+iโข๐’ฅ)โŸผโˆ’(Jx+iโขJy)Y=\begin{pmatrix}0&0\\ 1&0\end{pmatrix}=-(\mathcal{I}+i\mathcal{J})\longmapsto-(J_{x}+iJ_{y})

and similarly Xโ†ฆJxโˆ’iโขJyX\mapsto J_{x}-iJ_{y}.

Corollary 3.15.6.

For each โ„“โ‰ฅ0\ell\geq 0, there is a unique irreducible representation V(โ„“)V^{(\ell)} of SOโข(3)\mathrm{SO}(3) of dimension 2โขโ„“+12\ell+1 with derivative isomorphic to the representation Sym2โขโ„“โก(V)\operatorname{Sym}^{2\ell}(V) of ๐”ฐโข๐”ฌ3\mathfrak{so}_{3}. This gives the complete list of irreducible representations of SOโข(3)\mathrm{SO}(3) up to isomorphism.

Proof.

We simply have to work out which irreducible representations Symkโก(V)\operatorname{Sym}^{k}(V) of ๐”ฐโข๐”ฌ3\mathfrak{so}_{3} exponentiate to a representation of SOโข(3)\mathrm{SO}(3). Since we have SUโข(2)/{ยฑId}โ‰…SOโข(3)\mathrm{SU}(2)/\{\pm\operatorname{Id}\}\cong\mathrm{SO}(3) and each Symkโก(V)\operatorname{Sym}^{k}(V) exponentiates to a unique representation of SUโข(2)\mathrm{SU}(2) โ€” which we also call Symkโก(V)\operatorname{Sym}^{k}(V) โ€” this is equivalent to asking for which kk the centre {ยฑId}\{\pm\operatorname{Id}\} of SUโข(2)\mathrm{SU}(2) acts trivially on Symkโก(V)\operatorname{Sym}^{k}(V). But we see that โˆ’Id-\operatorname{Id} acts as (โˆ’1)k(-1)^{k}, so the answer is for even kk only.

Thus Symkโก(V)\operatorname{Sym}^{k}(V) exponentiates to a representation of SOโข(3)\mathrm{SO}(3) if, and only if, k=2โขโ„“k=2\ell is even, and we obtain the result. โˆŽ

We can consider the weights of these representations. Under the isomorphism

๐”ฐโข๐”ฒ2,โ„‚=๐”ฐโข๐”ฉ2,โ„‚โŸถ๐”ฐโข๐”ฌ3,โ„‚\mathfrak{su}_{2,\mathbb{C}}=\mathfrak{sl}_{2,\mathbb{C}}\longrightarrow% \mathfrak{so}_{3,\mathbb{C}}

the element H=โˆ’2โขiโข๐’ฆH=-2i\mathcal{K} maps to โˆ’2โขiโขJz-2iJ_{z}. Since the HH-weights of V(โ„“)V^{(\ell)} are โˆ’2โขโ„“,โˆ’2โข(โ„“โˆ’1),โ€ฆ,2โข(โ„“โˆ’1),2โขโ„“-2\ell,-2(\ell-1),\ldots,2(\ell-1),2\ell, we must divide these by โˆ’2โขi-2i to find the weights of JzJ_{z} acting on V(โ„“)V^{(\ell)}:

โˆ’iโขโ„“,โˆ’iโข(โ„“โˆ’1),โ€ฆ,iโข(โ„“โˆ’1),iโขโ„“.-i\ell,-i(\ell-1),\ldots,i(\ell-1),i\ell.
Example 3.15.7.

Consider the standard three-dimensional representation of SOโข(3)\mathrm{SO}(3) on โ„‚3\mathbb{C}^{3}. The weights of JzJ_{z} are simply its eigenvalues as a 3ร—33\times 3 matrix, which are โˆ’i,0,i-i,0,i. We see that this representation is isomorphic to V(1)V^{(1)}.

3.15.3. Exercises

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Problemย 40.

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    For aโ‰ฅba\geq b integers, decompose the representation Symaโกโ„‚2โŠ—Symbโกโ„‚2\operatorname{Sym}^{a}\mathbb{C}^{2}\otimes\operatorname{Sym}^{b}\mathbb{C}^{2} of ๐”ฐโข๐”ฉ2,โ„‚\mathfrak{sl}_{2,\mathbb{C}} into irreducibles. (This is known as the Clebschโ€“Gordan formula).

  2. (b)
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    (+) Can you find a general expression for the highest weight vectors for the irreducible subrepresentations? What about for the weight bases?