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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?
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Related: 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

I) Representation theory for the restricted$^1$ Lorentz group $SO^+(1,3;\mathbb{R})\cong SL(2,\mathbb{C})/\mathbb{Z}_2$ is a fairly broad subject covered in many textbooks, see e.g. Ref. 1 for further information.

An irreducible representation

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

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.


  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 symmetrized 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 $SO(3,1;\mathbb{C})$. One can show that the double cover of the complexified Lorentz group $SO(3,1;\mathbb{C})$ 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

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

in (1) of the product Lie group (4) is given as

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


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

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


  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 $O(3,1;\mathbb{R})$.

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

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Thanks for your answer. Your write: $\tag{5} \rho_{L/R}:SL(2,\mathbb{C})\to GL(V_{L/R},\mathbb{C})$, which I do agree with. Unfortunately I still can't see the answer to my original question: What is the vector space $V=V_L\otimes V_R$ here, for the different irreducible reprsentations recited above? – Tim Dec 3 '14 at 8:41
I updated the answer. – Qmechanic Dec 3 '14 at 9:34
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


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


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


$(0,\frac{1}{2})$ 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 >+\delta | \Downarrow> ) $


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


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

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


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

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

I am using $\oplus$, since $<s_1,p_i|s_2,p_j>=\delta_{ij} <s_1 |s_2 >$

Therefore the space is: $$(\sum_{s=1,2} \sum_{p} a_{s,p} |s,p>)$$ 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> \} $ itself

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