# Vector spaces for the irreducible representations of the Lorentz Group

EDIT: The vector space for the $(\frac{1}{2},0)$ Representation is $\mathbb{C}^2$ as mentioned by Qmechanic in the comments to his answer below! The vector spaces for the other representations remain unanswered.

The definition of a representation is a map (a homomorphism) to the space of linear operators over a vector space. My question is: What are the corresponding vector spaces for the

• $(0,0)$ Representation
• $(\frac{1}{2},0)$ Representation
• $(0,\frac{1}{2})$ Representation

• $(\frac{1}{2},0) \oplus (0,\frac{1}{2})$ Representation

• $(\frac{1}{2},\frac{1}{2})$ Representation

• infinite dimensional Representation?
• Related: physics.stackexchange.com/q/28505/2451 and links therein. – Qmechanic Nov 29 '14 at 12:54
• @Qmechanic After reading through your answer, I'm still not sure. Is the corresponding vectorspace for the $(\frac{1}{2},0)$ representation the space of complex $2\times 2$ matrices ${\rm Mat}_{2\times 2}$? In what sense is the vector space for the $(0,\frac{1}{2})$ representation different? For the $(\frac{1}{2},\frac{1}{2})$ Representation we take the product space and for $(\frac{1}{2},0) \oplus (0,\frac{1}{2})$ the direct sum of those two vector spaces?! – Tim Dec 1 '14 at 9:31
• This sounds like a rather confused question. The vector space for any $n$-dimensional complex representation is always $\mathbb{C}^n$. Depending on the physical context, for a particular representation you might give the basis vectors special names (such as "spin up" and "spin down"), but the vector space itself is just $\mathbb{C}^n$. – knzhou Nov 20 '18 at 13:45

## 2 Answers

I) Representation theory for the double cover $$SL(2,\mathbb{C})$$ of the restricted$$^1$$ Lorentz group $$SO^+(1,3;\mathbb{R})$$ is a fairly broad subject covered in many textbooks, see e.g. Ref. 1 for further information.

$$\tag{1} (j_L,j_R)~=~j_L\otimes_{\mathbb{C}} j_R, \qquad j_L, j_R~\in~ \frac{1}{2}\mathbb{N}_0,$$

is a tensor product of $$V=V_L\otimes_{\mathbb{C}} V_R$$ of two complex vector spaces $$V_L$$ and $$V_R$$, of complex dimension $$2j_L+1$$ and $$2j_R+1$$, respectively. The tensor product $$V$$ is again a complex vector space and has complex dimension $$(2j_L+1)(2j_R+1)$$. See also this Phys.SE post.

Examples:

1. $$(j_L,j_R)=(0,0)$$. This is the trivial/singlet representation. Then the vector-space is $$V\cong\mathbb{C}$$. Note that the trivial representation $$(0,0)$$ is the multiplicative identity for the tensor product $$\otimes_{\mathbb{C}}$$, i.e. $$\tag{2}\forall V:~~(0,0)\otimes_{\mathbb{C}}V~\cong~ V~\cong~ V\otimes_{\mathbb{C}}(0,0).$$

2. $$(j_L,j_R)=(\frac{1}{2},0)$$. This is known as the left-handed Weyl-spinor representation. Then the vector-space is $$V\cong\mathbb{C}^2$$. It is the fundamental/defining representation of $$SL(2,\mathbb{C})$$.

3. $$(j_L,j_R)=(0,\frac{1}{2})$$. This is known as the right-handed Weyl-spinor representation. It is the complex conjugate representation of the left-handed Weyl-spinor representation.

An irreducible representation (1) can be written with the help of the symmetric tensor product $$\odot$$ of the left-handed and right-handed Weyl-spinor representation

$$(j_L,j_R)~=~(\frac{1}{2},0)^{\odot 2j_L} \otimes (0,\frac{1}{2})^{\odot 2j_R}$$ $$~:=~\underbrace{\left\{(\frac{1}{2},0)\odot\ldots\odot(\frac{1}{2},0)\right\}}_{2j_L\text{ symmetrized factors}} \otimes \underbrace{\left\{(0,\frac{1}{2})\odot\ldots\odot(0,\frac{1}{2})\right\}}_{2j_R\text{ symmetrized factors}} .\tag{3}$$

Here $$\otimes$$ denotes the standard (un-symmetrized) tensor product.

II) Complexification. The restricted Lorentz group $$SO^+(1,3;\mathbb{R})$$ is obviously a subgroup of the complexified$$^2$$ Lorentz group $$SO(1,3;\mathbb{C})$$. One can show that the double cover of the complexified Lorentz group $$SO(1,3;\mathbb{C})$$ is isomorphic to the direct or Cartesian product group

$$\tag{4} G~=~SL(2,\mathbb{C})_L\times SL(2,\mathbb{C})_R,$$

cf. e.g. Ref. 1 and this Phys.SE post.

In more detail, the irreducible representation (1) for $$SL(2,\mathbb{C})$$ lifts to an irreducible representation

$$\tag{5} \rho~=~\rho_L\otimes \rho_R:G\to GL(V,\mathbb{C})$$

for the product Lie group (4) given as

$$\tag{6} \rho(g_L,g_R)(\sum_iv^i_L\otimes v^i_R)~=~\sum_i\rho_L(g_L)v^i_L\otimes\rho_R(g_R)v^i_R ,$$

where both

$$\tag{7} \rho_{L/R}:SL(2,\mathbb{C})\to GL(V_{L/R},\mathbb{C})$$

are irreducible representations of $$SL(2,\mathbb{C})$$ of complex dimensions $$2j_{L/R}+1$$.

References:

1. I.L. Buchbinder and S.M. Kuzenko, Ideas and Methods of Supersymmetry and Supergravity - Or a Walk Through Superspace, 1998; Chapter 1.

--

$$^1$$ Let us here for simplicity consider the restricted Lorentz group $$SO^+(1,3;\mathbb{R})$$ rather than the Lorentz group $$O(1,3;\mathbb{R})$$. To allow for spinor representations, we need to go to the double cover $$SL(2,\mathbb{C})$$.

$$^2$$ It turns out that relativistic physical theories often have pertinent complex analytic properties.

• Thanks for your help. Unfortunately I'm still can't find the answer to my question in your answer. I even went to the local libary to get the book you recommended, but couldn't find the answer. I think my question is much more simpler than what your answer is aiming for. I'm trying to understand what the vector space $V$ in $\tag{3} \rho~=~\rho_L\otimes \rho_R:G\to GL(V,\mathbb{C})$ explicitly (not genreally) is? – Tim Jan 20 '15 at 8:09
• Example: $(j_L,j_R)=(\frac{1}{2},0)$. This is known as the left-handed Weyl-spinor representation. Then the vector-space is $V\cong\mathbb{C}^2$. – Qmechanic Jan 20 '15 at 8:19
• Thank you so much! This is exactly the kind of answer I'm looking for. Is for the right-handed Weyl spinor rep $(0,\frac{1}{2})$ the vector space $V\cong\mathbb{C}^2$, too? What's the vector space for the $(\frac{1}{2},\frac{1}{2})$ rep? ( I know this is the vector representations and the objects transform like the usual four-vectors living in Minkowski space. Nevertheless, naturally we have that this representation acts on hermitian $2 \times 2$ matrices. What's the name for the vector space they live in? Is it something like $\mathbb{C}^2 \otimes \mathbb{C}^2)$? ) – Tim Jan 20 '15 at 8:31

(A)

(0,0) acts on a trivial space $$\mathbb{C}.$$

(B)

$$(\frac{1}{2},0)$$ acts on a vector space which is same as a spin space $$( \alpha|\uparrow \rangle +\beta | \downarrow\rangle)$$ , ignoring the meaning of spin up and down now. This space is just $$\mathbb{C}^2$$ up to a normalization constraint $$|\alpha|^2+|\beta|^2=1.$$

(C)

$$(0,\frac{1}{2})$$ acts on a vector space, which has the same structure as $$(\frac{1}{2},0)$$'s space, but may have different meaning, I write it as $$( \gamma|\Uparrow \rangle +\delta | \Downarrow\rangle).$$

(D)

$$(\frac{1}{2},0) \oplus (0,\frac{1}{2})$$ acts on $$(\alpha|\uparrow \rangle +\beta | \downarrow\rangle) \oplus (\gamma|\Uparrow\rangle +\delta | \Downarrow\rangle)=( \alpha|\uparrow\rangle +\beta |\downarrow\rangle + \gamma|\Uparrow\rangle +\delta | \Downarrow\rangle).$$

(E)

$$(\frac{1}{2}, \frac{1}{2})$$ acts on $$(\alpha|\uparrow\rangle +\beta | \downarrow\rangle)\otimes (\gamma|\Uparrow\rangle +\delta | \Downarrow\rangle )=(a|A\rangle + b|B\rangle +c|C\rangle +d|D\rangle).$$

$$|\alpha|^2+|\beta|^2=1$$ and $$|\gamma|^2+|\delta|^2=1$$ may not hold, it becomes one expression for $$a \ b \ c \ d.$$

(F)

infinite basis, adding extra momentum to (B) for example:

$$(\alpha_1|\uparrow,p_1\rangle +\beta_1 | \downarrow,p_1\rangle)\oplus( \alpha_2|\uparrow,p_2\rangle +\beta_2 | \downarrow,p_2\rangle)\oplus( \alpha_3|\uparrow,p_3 \rangle +\beta_3 | \downarrow,p_3\rangle)\oplus...$$

I am using $$\oplus$$, since $$\langle s_1,p_i|s_2,p_j\rangle =\delta_{ij} \langle s_1 |s_2\rangle .$$

Therefore the space is:

$$(\sum_{s=1,2} \sum_{p} a_{s,p} |s,p\rangle)$$ with normalization constraint $$\sum_{s=1,2} \sum_{ p} |a_{s,p}|^2=1.$$

similarly, you can add extra momentum to (A) (C) (D) (E), to realize their infinite versions.

for (A)'s infinite version, that vector space is just $$\{ |p\rangle \}$$ itself.

• (D) is not an irreducible representation, is it? – gented Nov 9 '16 at 0:27
• @gented . No, the bispinor rep is reducible, cf. Dirac spinors. – Cosmas Zachos May 2 '18 at 21:38