What is SR transformarion for bispinor?

[olgerm]

what is SR transformarion for bispinor? I have heard that it is different than 4-vectors transfomation.

 

[king vitamin]

The Wikipedia article gives a pretty nice overview of how Lorentz transformations act on bispinors: https://en.wikipedia.org/wiki/Bispinor. Perhaps you can read through that and then ask more specific questions on what you don't understand there.

In particular, SO(3,1) transformations are block diagonal in the Weyl basis, which is perhaps the most important property of this representation (and is not true for the vector representation which is also four-dimensional).

 

[olgerm]

The Wikipedia article gives a pretty nice overview of how Lorentz transformations act on bispinors: https://en.wikipedia.org/wiki/Bispinor

Yes.
$\chi=v/c$?

 

[olgerm]

Yes.
$\chi=v/c$?

It is possible to write that as one (4,4) matrix. So that it does not include matrix exponentation.

 

[king vitamin]

Yes.
χ=v/c?

Good catch, it seems that Wikipedia is using notation from these lecture notes.
In particular, for a boost with velocity $v_i/c$ in the $i$th direction, the "boost parameter" is the so-called rapidity:
$$
v_i/c = \tanh \chi_i
$$
The point of introducing this quantity is that it appears in boosts in an analogous way to how the angle appears in rotations (but with hyperbolic instead of trigonometric functions, see the linked Wikipedia article). So if you want to boost by a general velocity vector $\vec{v}$, you should form the corresponding vector of rapidities $\vec{\chi}$ to use in that formula.

It is possible to write that as one (4,4) matrix. So that it does not include matrix exponentation.

Yes, one simply uses the following formula, which you should attempt to prove (it is a good exercise for somebody working on this stuff):
$$
e^{a \, \hat{n} \cdot \vec{\sigma}} = \mathbf{I} + \hat{n} \cdot \vec{\sigma} \sinh a .
$$
Here, I have written the vector in the exponential as a magnitude times a unit vector: $\vec{\chi} = a \, \hat{n}$ where $\hat{n} \cdot \hat{n} = 1$.

 

[olgerm]

and is not true for the vector representation which is also four-dimensional)

Can you send me a link or explain what you meant by vector representation?

 

[king vitamin]

Can you send me a link or explain what you meant by vector representation?

The vector representation of the Lorentz group is also called the "fundamental" representation, as it is the usual 4x4 matrix representation of Lorentz transformations acting on four-vectors. A boost in the vector representation is given by this matrix. Here, the boost is in the direction $\hat{n} = (n_x,n_y,n_z)$ (where $\hat{n} \cdot \hat{n} = 1$), and $\beta = v/c$, $\gamma = 1/\sqrt{1 - (v/c)^2}$. See also the info here. As you can see, it is not block diagonal as the bispinor boosts are.

Given that you're asking this question, I think maybe we should back up. What prior knowledge are you coming to these questions with? In your original post you say you had heard that the bispinor rep "is different than 4-vectors transfomation," yet now you are asking what the vector representation is. Do you have any textbooks which deal with the representation theory of the Lorentz group at any level, and if so, what is it? If you are coming here with a small background on representation theory then my answers have not been at the appropriate level, so it'd be more useful for you if I knew your level.

 

[Demystifier]

bispinors: https://en.wikipedia.org/wiki/Bispinor.

According to this, bispinor is a direct sum of two spinors. I thought that bispinor should be a direct product of two spinors, in which case it would be equivalent to a vector.

 

[haushofer]

According to this, bispinor is a direct sum of two spinors. I thought that bispinor should be a direct product of two spinors, in which case it would be equivalent to a vector.

I'm also used to that last interpretation in the context of supergravity, e.g. in the susy-transfo of vielbeine.

 

[vanhees71]

According to this, bispinor is a direct sum of two spinors. I thought that bispinor should be a direct product of two spinors, in which case it would be equivalent to a vector.

A Dirac spinor is the direct sum of two different Weyl spinors, one transforming according to the (1/2,0) the other according to the (0,1/2) representation of the proper orthochronous Lorentz group. You need both spinors to also enable the space-reflection symmetry (parity conservation) of electrodynamics and QCD, i.e., the Dirac spinor refers to the representation $(1/2,0) \oplus (0,1/2)$ (which is reducible for the proper orthochronous Lorentz group but irreducible as a representation of the orthchronous Lorentz group, which additionally includes space reflections).

In contradistinction to this the vector representation of the proper orthochronous Lorentz group is the irreducible representation $(1/2,1/2)$ of the proper orthochronous Lorentz group.

 

[martinbn]

According to this, bispinor is a direct sum of two spinors. I thought that bispinor should be a direct product of two spinors, in which case it would be equivalent to a vector.

There is no difference in this case.

 

[vanhees71]

There's a difference between $(1/2,0) \oplus (0,1/2)$ and $(1/2,1/2)=(1/2,0) \otimes (0,1/2)$.

 

[martinbn]

There's a difference between $(1/2,0) \oplus (0,1/2)$ and $(1/2,1/2)=(1/2,0) \otimes (0,1/2)$.

There is no difference between the direct sum and direct product of two vector spaces or any finite number of vector spaces. You need an infinite number of them to see a difference (not to be confused with the dimensions of the vector spaces, that makes no difference). Check the definitions.

What you wrote is unclear to me. You use the tensor product notation when you talk about the direct product. Also $(1/2,1/2)$ is very strange, to say the least, it suggests two dimensions, while the space is four dimensional!

 

[king vitamin]

According to this, bispinor is a direct sum of two spinors. I thought that bispinor should be a direct product of two spinors, in which case it would be equivalent to a vector.

I'm also used to that last interpretation in the context of supergravity, e.g. in the susy-transfo of vielbeine.

Interesting, I admit that I wasn't familiar with the term, so I just googled it and clicked on the Wikipedia article. I would usual call this object a "Dirac spinor," but it seems that Wikipedia reserves this term for bispinors which are specifically plane wave solutions of the Dirac equation.

Doing some more googling of the term "bispinor," it seems that the Wikipedia definition is fairly common, but if I include the search term "supergravity" as well, then the definition you are both familiar with comes up. What an annoying clash of terminology. Hopefully I answered the question which OP was actually interested in.

There is no difference between the direct sum and direct product of two vector spaces or any finite number of vector spaces.

If I have two finite-dimensional vector spaces with dimension $d_1$ and $d_2$, then the direct sum of the two vector spaces has dimension $d_1 + d_2$, while the direct product has dimension $d_1 d_2$. Therefore, they cannot be equivalent in general. (This particular argument isn't so great for the example in the above post where $d_1 = d_2 = 2$, but it clearly shows that the quoted statement is incorrect.)

Also (1/2,1/2) is very strange, to say the least, it suggests two dimensions, while the space is four dimensional!

I think you are just unfamiliar with the notation being used. The dimension of the $(s,s')$ irrep of the Lorentz group has dimension $(2s+1)(2s'+1)$, so this irrep is four-dimensional. (As @vanhees71 claimed, it is the four-vector irrep.)

 

[martinbn]

If I have two finite-dimensional vector spaces with dimension $d_1$ and $d_2$, then the direct sum of the two vector spaces has dimension $d_1 + d_2$, while the direct product has dimension $d_1 d_2$. Therefore, they cannot be equivalent in general. (This particular argument isn't so great for the example in the above post where $d_1 = d_2 = 2$, but it clearly shows that the quoted statement is incorrect.)

No, the direct product has dimension $d_1 + d_2$. Look up the definition.

I think you are just unfamiliar with the notation being used. The dimension of the $(s,s')$ irrep of the Lorentz group has dimension $(2s+1)(2s'+1)$, so this irrep is four-dimensional. (As @vanhees71 claimed, it is the four-vector irrep.)

I see, the notations are for the representation spaces, not vectors in them. Then he definitely means tensor product, not direct product.

 

[king vitamin]

I see, the notations are for the representation spaces, not vectors in them. Then he definitely means tensor product, not direct product.

Yes, a second case of unfortunate terminology in this thread. Certain physics textbooks use the terms direct product and tensor product interchangeably when referring to representations (which I imagine drives mathematicians nuts). I think that everyone above meant tensor product (I did), but they'd have to clarify for themselves.

 

[George Jones]

I think that everyone above meant tensor product (I did), but they'd have to clarify for themselves.

The original post refers to a direct sum (or direct product) of (inequivalent) representations, not to a tensor product of representations.

A Dirac spinor is the direct sum of two different Weyl spinors, one transforming according to the (1/2,0) the other according to the (0,1/2) representation of the proper orthochronous Lorentz group. You need both spinors to also enable the space-reflection symmetry (parity conservation) of electrodynamics and QCD, i.e., the Dirac spinor refers to the representation $(1/2,0) \oplus (0,1/2)$ (which is reducible for the proper orthochronous Lorentz group but irreducible as a representation of the orthchronous Lorentz group, which additionally includes space reflections).

 

[king vitamin]

The original post refers to a direct sum (or direct product) of (inequivalent) representations, not to a tensor product of representations.

Right, hence why my original reply referred to the representation being block diagonal. What I meant in your quote (which I hope is clear from the context) is that I was later using the word "direct product" interchangeably with "tensor product" in my post #14. This is probably bad practice from a mathematical POV, but in my defense, many textbooks (Merzbacher, Weinberg, Zee...) do the same, so I also do so sometimes.

 

[Demystifier]

Then he definitely means tensor product, not direct product.

It seems that here we have a semantical confusion, because what mathematicians call tensor product, physicists often call direct product. See e.g. https://www.quora.com/What-is-the-m...-the-context-of-physics-and-quantum-mechanics

 

[Demystifier]

The original post refers to a direct sum (or direct product) of (inequivalent) representations, not to a tensor product of representations.

In your dictionary, what would be the difference between direct product and tensor product?

 

[Demystifier]

After some googling, I think I understand now where the confusion and distortion of terminology came from. In mathematics of vector spaces https://en.wikipedia.org/wiki/Tensor_product we have two closely related notions, namely tensor product for which
$${\rm dim}(V\otimes W)={\rm dim}V\times {\rm dim}W$$
and direct sum for which
$${\rm dim}(V\oplus W)={\rm dim}V+ {\rm dim}W$$
But since the latter is called direct sum, and not tensor sum, it seems natural to call the former direct product. That must be how some physicists started to actually call the former direct product instead of tensor product.

 

[George Jones]

Right, hence why my original reply referred to the representation being block diagonal. What I meant in your quote (which I hope is clear from the context) is that I was later using the word "direct product" interchangeably with "tensor product" in my post #14.

Sorry, for me, it sometime takes quite a while for the penny to drop. In this case, this happened some time after I posted.

This is probably bad practice from a mathematical POV, but in my defense, many textbooks (Merzbacher, Weinberg, Zee...) do the same, so I also do so sometimes.

As do I, but see below for a clash of terminology, even for physicists, that only occurred to me this morning. (An example of an even more slowly dropping penny.)

In your dictionary, what would be the difference between direct product and tensor product?

Which dictionary, my dictionary of terms used by physicists, or my dictionary of terms used by mathematicians? Both types of creature lurk along my hall, so, in attempt at self-preservation, I have both dictionaries.

As I was walking to my office this morning, I thought about some motivation for the direct sum-like definition of "direct product of vector spaces", which involves things that only seem to be unrelated to the topic.

Let $G$ and $H$ be groups. The direct product of $G$ and $H$ is the set $G \times H$, together with a group product on $G \times H$ defined by
$$\left( g_1 , h_1 \right) \left( g_2 , h_2 \right) := \left( g_1 g_2 , h_1 h_2 \right). $$

This definition "direct product of groups" is used both in theoretical physics references, and in pure math mathematics references.

When "+" and "0" occur in some mathematical structure, it usually means that the part of the structure involves an abelian group, with "+" denoting the group product, and "0" denoting the identity element of the group. This is true for vector spaces. In turn, this means that the above definition of group direct product (used by both physicists and mathematicians) applies to vector spaces.

For example, if $V$ and $W$ are vector spaces, then the above definition of direct product of groups (with group product denoted by "+" instead of juxtaposition) together with an obvious definition of scalar multiplication results in a definition of direct product of vector space that looks much like a direct sum.

In other categories, direct sums and direct product can look much different.

 

[formodular]

A four vector $x^{\mu}$ transforms under a Lorentz transformation as $x^{\mu} \to x'^{\mu} = (e^{-\frac{i}{2}\omega_{\rho \sigma} J^{\rho \sigma}})^{\mu} \, _{\nu} x^{\nu}$ where $J^{\rho \sigma}$ generate the vector representation of the Lorentz algebra, $(J^{\rho \sigma})^{\mu} \, _{\nu} = i (\eta^{\mu \rho}\delta^{\sigma} \, _{\nu} - \eta^{\mu\sigma}\delta^{\rho} \, _{\nu})$, which can be derived, for example, from the infinitesimal Lorentz transformation of a four-vector $x'^{\mu} = \Lambda^{\mu} \, _{\nu} x^{\nu}$, i.e. $x'^{\mu} = \Lambda^{\mu} \, _{\nu} x^{\nu} = (\delta^{\mu} \, _{\nu} + \omega^{\mu} \, _{\nu}) x^{\nu} = x^{\mu} - \frac{i}{2} \omega_{\rho \sigma} i (\eta^{\mu \rho}\delta^{\sigma} \, _{\nu} - \eta^{\mu\sigma}\delta^{\rho} \, _{\nu}) x^{\nu}$.

A bispinor transforms under a Lorentz transformation as $\psi^{\mu} \to \psi'^{\mu} = (e^{-\frac{i}{2}\omega_{\rho \sigma} J^{\rho \sigma}})^{\mu} \, _{\nu} \psi^{\nu}$ where $J^{\rho \sigma}$ generate the bi-spinor representation of the Lorentz algebra, $(J^{\rho \sigma})^{\mu} \, _{\nu} = \frac{i}{4} [\gamma^{\rho},\gamma^{\sigma}]^{\mu} \, _{\nu}$, where the $\gamma^{\mu}$ are the gamma matrices. Usually the bi-spinor representation is denoted by something like $S^{\rho \sigma}$ rather than $J^{\rho \sigma}$.

 

[martinbn]

In other categories, direct sums and direct product can look much different.

Even in the category of vector spaces the direct product and direct sum of an infinite family are different. For example the direct product of $\{V_i\}_{i=1}^\infty$ is the vector space of all sequences $(v_1,v_2,\dots)$ with $v_i\in V_i$ and componentwise operations. The direct sum is the vector space of sequences with only finitely many nonzero terms.

 

[vanhees71]

There is no difference between the direct sum and direct product of two vector spaces or any finite number of vector spaces. You need an infinite number of them to see a difference (not to be confused with the dimensions of the vector spaces, that makes no difference). Check the definitions.

What you wrote is unclear to me. You use the tensor product notation when you talk about the direct product. Also $(1/2,1/2)$ is very strange, to say the least, it suggests two dimensions, while the space is four dimensional!

We are talking about the representations of the Lorentz group not simply about direct sums/products of vector spaces. The Dirac-spinor representation of the covering group of the orthochronous Lorentz group (related to particles with spin 1/2) is definitely different and non-equivalent from the "fundamental representation" of $\mathrm{SO}(1,3)$ (related to particles with spin 1).

The notation $(j_1,j_2)$ refers to the treatment of the representation theory in terms of the two independent (complexified) "angular-momentum algebras" making up the Lie algebra $\mathfrak{sl}(2,\mathbb{C})$. For details, see Appendix B of

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf

 

[olgerm]

Yes, one simply uses the following formula, which you should attempt to prove (it is a good exercise for somebody working on this stuff)

it should be
\begin{smallmatrix}
\frac{1}{2}*e^{1 - e^{\frac{1}{2}*arctanh(v)}} + \frac{1}{2}*e^{1 + e^{\frac{1}{2}*arctanh(v)}} & -\frac{1}{2}*e^{1 - e^{\frac{1}{2}* arctanh(v)}} + \frac{1}{2}*e^{1 + e^{\frac{1}{2}*arctanh(v)}} & 0 & 0\\
-\frac{1}{2}*e^{1 - e^{1/2 arctanh(v)}} + \frac{1}{2}*e^{1 + e^{\frac{1}{2}*arctanh(v)}} & \frac{1}{2}*e^{1 - e^{\frac{1}{2}* arctanh(v)}} + \frac{1}{2}*e^{1 + e^{\frac{1}{2}*arctanh(v)}}& 0 & 0 \\
0 & 0 &\frac{1}{2}*e^{1 - e^{-\frac{1}{2}*arctanh(v)}} + \frac{1}{2}*e^{1 + e^{-\frac{1}{2}*arctanh(v)}} & -\frac{1}{2}*e^{1 - e^{-\frac{1}{2}* arctanh(v)}} + \frac{1}{2}*e^{1 + e^{-\frac{1}{2}*arctanh(v)}}\\
0 & 0 &-\frac{1}{2}*e^{1 - e^{-1/2 arctanh(v)}} + \frac{1}{2}*e^{1 + e^{-\frac{1}{2}*arctanh(v)}} & \frac{1}{2}*e^{1 - e^{-\frac{1}{2}* arctanh(v)}} + \frac{1}{2}*e^{1 + e^{-\frac{1}{2}*arctanh(v)}}
\end{smallmatrix}
Using units where c=1.
I used wolframalpha and wolframalpha to simplify it.

 

[olgerm]

king vitamin, I know what 4-vectors are and how they are transformed with poincare group transformations. I asked this question to find a way to describe particlefields that are usually described with bispinors with 4-vectors. Maybe this fundametnal description is what I have been looking for.
You are kind of right that I have to small preknowledge, but still try to answer my questions.

 

[samalkhaiat]

According to this, bispinor is a direct sum of two spinors.

In this instance and historically, bi-spinor refers to the 4-component spinor of Dirac $$\Psi_{D} = \begin{pmatrix} \psi_{\alpha} \\ \bar{\chi}^{\dot{\alpha}} \end{pmatrix} ,$$ or that of Majorana $$\Psi_{M} = \begin{pmatrix} \psi_{\alpha} \\ \bar{\psi}^{\dot{\alpha}} \end{pmatrix} , \ \ \alpha , \dot{\alpha} = 1,2 \ ,$$ where $\psi_{\alpha} \in (\frac{1}{2} , 0)$ is the 2-component (left-handed) Weyl spinor, and $\bar{\chi}_{\dot{\alpha}} \in (0 , \frac{1}{2})$ is the 2-component (right-handed) Weyl spinor, i.e., the two inequivalent fundamental representations of $\mbox{SL}(2 , \mathbb{C}) \cong \mbox{Spin}(1,3)$, the universal covering group of the Lorentz group $\mbox{SO}(1,3)^{\uparrow}$.

I thought that bispinor should be a direct product of two spinors, in which case it would be equivalent to a vector.

If I replace your “direct product” by tensor product, and your “vector” by Lorentz vector, your statement will still be only partially correct. This is obvious because you obtain spin-0 and spin-1 from the tensor product of two spin-1/2, and geometrically, spin-0 corresponds to a Lorentz scalar, and spin-1 corresponds either to a Lorentz vector, or to an anti-symmetric Lorentz tensor.

Higher rank spin tensors, say $S_{\alpha_{1} \cdots \alpha_{n} \dot{\alpha}_{1} \cdots \dot{\alpha}_{m}}$ can be obtained by taking the tensor products of $n$-copy of $\psi_{\alpha} \in (\frac{1}{2} , 0)$ with $m$-copy of $\bar{\chi}_{\dot{\alpha}} \in (0 , \frac{1}{2})$. In particular, we can form 3 different rank-2 spin tensors: $S_{\alpha \dot{\alpha}} \equiv \psi_{\alpha}\bar{\chi}_{\dot{\alpha}}$ , $S_{\alpha \beta} \equiv \psi_{\alpha}\chi_{\beta}$ and $S_{\dot{\alpha}\dot{\beta}}$. Few people (only few) attached the name “bi-spinors” to these rank-2 spin tensors $S$.
Using the properties of the $\sigma_{a}$-matrices: $(\sigma_{a})_{\alpha \dot{\alpha}}$ and $(\bar{\sigma}_{a})^{\dot{\alpha}\alpha}$, and the generators $(\sigma_{ab})_{\alpha\beta}$ and $(\bar{\sigma}_{ab})_{\dot{\alpha}\dot{\beta}}$, one can make a complete dictionary between spin tensors and Lorentz tensors. For example, the spin tensor $S_{\alpha \dot{\alpha}}$ corresponds to a Lorentz vector. Indeed, one finds $$S_{\alpha \dot{\alpha}} = - \frac{1}{2} (\sigma_{a})_{\alpha \dot{\alpha}} V^{a}, \ \ \ \ \ \ \ (1)$$ where $V^{a} = \psi \sigma^{a} \bar{\chi}$ transforms as a vector with respect to the Lorentz group $\mbox{SO}(1,3)^{\uparrow}$. One can also show that the spin tensor $S_{\alpha \beta}$ corresponds to a Lorentz scalar and anti-symmetric Lorentz tensor. In fact $$S_{\alpha \beta} = \frac{1}{2} \epsilon_{\alpha \beta} \Phi - \frac{1}{2} (\sigma^{ab})_{\alpha \beta} F_{ab} , \ \ \ \ \ (2)$$ where $\Phi = \psi^{\eta}\chi_{\eta}$ is a Lorentz scalar, and $F_{ab} = \psi \sigma_{ab}\chi$ is an anti-symmetric Lorentz tensor. In group theoretical language, Eq(2) means that $(\frac{1}{2} , 0) \otimes (\frac{1}{2} , 0) = (0,0) \oplus (1 , 0)$, while Eq(1) means that $(\frac{1}{2} , 0) \otimes (0 , \frac{1}{2}) = (\frac{1}{2} , \frac{1}{2})$.

 

[PeterDonis]

spin-1 corresponds either to a Lorentz vector, or to an anti-symmetric Lorentz tensor.

Is this correct? A 4-vector has 4 components, but an antisymmetric 4-tensor has 6 independent components.

 

[samalkhaiat]

Is this correct? A 4-vector has 4 components, but an antisymmetric 4-tensor has 6 independent components.

Yes, think of the vector potential and the field tensor of Maxwell. Also, the spin of a Lorentz vector (1/2,1/2) is 1/2+1/2 = 1and the number of components is equal to the dimention of (1/2,1/2) which is [2(1/2) +1][2(1/2) +1]=4. Now, the spin of the anti-symmetric tensor (1,0) is equal to 0+1=1 and the number of complex components is equal to [2(1) + 1][2(0) + 1] = 3.

 

[PeterDonis]

think of the vector potential and the field tensor of Maxwell

The fact that one particular antisymmetric 4-tensor is the exterior derivative of a 4-vector (which I assume is why you're saying that in this case there are really only 4 independent components, not 6) does not mean every antisymmetric 4-tensor is the exterior derivative of a 4-vector.

the number of complex components is equal to [2(1) + 1][2(0) + 1] = 3.

Which is 6 real components, not 4.

 

[samalkhaiat]

Which is 6 real components, not 4.

Why do you need them to be 4?I gave you the rule to calculate the spin and dimention of a representation. The vector representation (1/2,1/2) is real, while the tensor (1,0) is complex. For Maxwell field tensor, we associate the real representation (1,0) + (0,1).

 

[PeterDonis]

Why do you need them to be 4?

I don't. I'm trying to understand how spin-1 can "correspond to an antisymmetric Lorentz tensor", as you said, when "spin-1", to me, means 4 independent components, but "antisymmetric Lorentz tensor" means 6. Basically what I understand you to be saying is that my interpretation of "spin-1" is wrong--"spin-1" is a more general term that includes multiple representations, only one of which (the Lorentz vector) has 4 independent real components.

I gave you the rule to calculate the spin and dimention of a representation. The vector representation (1/2,1/2) is real, while the tensor (1,0) is complex. For Maxwell field tensor, we associate the real representation (1,0) + (0,1).

So by the rule you gave, I get that (1/2, 1/2) is spin 1 and (1, 0) is spin 1. Would (1, 0) + (0, 1) still be considered spin 1?

 

[samalkhaiat]

I don't. I'm trying to understand how spin-1 can "correspond to an antisymmetric Lorentz tensor", as you said, when "spin-1", to me, means 4 independent components, but "antisymmetric Lorentz tensor" means 6. Basically what I understand you to be saying is that my interpretation of "spin-1" is wrong--"spin-1" is a more general term that includes multiple representations, only one of which (the Lorentz vector) has 4 independent real components.

An irreducible representation $(j,k)$ has 2, in general, unrelated numbers. For non-zero mass, these are the spin and dimension: $$\mbox{Spin}(j,k) = j + k ,$$$$\mbox{dim}(j,k) = (2j +1)(2k +1).$$ The rule of identification with objects on Minkowski space-time is as follows: The irreducible representation $(j,k)$ corresponds to a massive spin-$(j + k)$ Lorentz tensor with $\mbox{dim}(j,k)$ independent components. For example the irrep $(1,1)$ corresponds to a (massive) spin-2, traceless symmetric Lorentz tensor: $G_{ab} = G_{ba}, \ \eta^{ab}G_{ab} = 0$. Check the rule: $\mbox{dim}(1,1) = 9$ and this is exactly the number of independent components in a traceless symmetric tensor in 4 dimensions.

Would (1, 0) + (0, 1) still be considered spin 1?

Yes $\mbox{Spin}(j,0) = \mbox{Spin}\left( (j , 0) \oplus (0 , j)\right)$

 

[vanhees71]

It's clear that you can represent fields of a given spin in different ways. E.g., you can represent a (massive) vector field by a four-vector field $A^{\mu}(x)$. Nothing else said, it has 4 field-degrees of freedom, but from representation theory we know that a spin-1 particle has only 3 spin (polarization) degrees of freedom. That's usually part of the equations of motion to project out the unwanted spin-degrees of freedom (in this case the spin-0 part). In both standard treatments (Proca or Stueckelberg) you end up with the contraint $\partial_{\mu} A^{\mu}=0$, which projects out the spin-0 part.

Another way is to use an antisymmetric tensor field with the constraint $\partial_{\mu} ^{\dagger} F^{\mu \nu}=0$.