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?
  • $\begingroup$ Related: physics.stackexchange.com/q/28505/2451 and links therein. $\endgroup$ – Qmechanic Nov 29 '14 at 12:54
  • $\begingroup$ @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?! $\endgroup$ – Tim Dec 1 '14 at 9:31
  • 3
    $\begingroup$ 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 $\mathbb{C}^n$. $\endgroup$ – knzhou Nov 20 '18 at 13:45

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.

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 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 $2j_{L/R}+1$.


  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.

  • 1
    $\begingroup$ 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? $\endgroup$ – Tim Jan 20 '15 at 8:09
  • 3
    $\begingroup$ Example: $(j_L,j_R)=(\frac{1}{2},0)$. This is known as the left-handed Weyl-spinor representation. Then the vector-space is $V\cong\mathbb{C}^2$. $\endgroup$ – Qmechanic Jan 20 '15 at 8:19
  • $\begingroup$ 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 $\mathbb{C}^2 \otimes \mathbb{C}^2)$? ) $\endgroup$ – 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 \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 $|\alpha|^2+|\beta|^2=1.$


$(0,\frac{1}{2})$ 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). $


$(\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). $


$(\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 $a \ b \ c \ d.$


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.

  • 1
    $\begingroup$ (D) is not an irreducible representation, is it? $\endgroup$ – gented Nov 9 '16 at 0:27
  • $\begingroup$ @gented . No, the bispinor rep is reducible, cf. Dirac spinors. $\endgroup$ – Cosmas Zachos May 2 '18 at 21:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.