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On the Wikipedia article about the representation theory of the Lorentz group, the finite-dimensional representations $(1,0)$ and $(0,1)$ are referred to as "$2$-form" representations. On the other hand, Carroll's notes say that $p$-forms are $(0,p)$ tensors. Is one of these incorrect, or does the word "form" have a different meaning in each case?

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Finite dimensional representations of the universal cover of the Lorentz group can be found by harnessing the complexified Lie algebra split ${\frak so}_{\mathbb{C}}(1,3)\simeq \mathfrak{su}_{\mathbb{C}}(2)\oplus \mathfrak{su}_{\mathbb{C}}(2)$. In particular, what this means is that after complexification, representations of ${\frak so}_{\mathbb{C}}(1,3)$ can be classified in terms of two representations of ${\frak su}_{\mathbb{C}}(2)$. This is very well explained in Weinberg's The Quantum Theory of Fields, Chapter 5. See also this Phys.SE thread.

Representations of ${\frak su}_{\mathbb{C}}(2)$, on the other hand, are well-known from the theory of angular momentum in non-relativistic Quantum Mechanics. Indeed, while most of the time this is not explicitly stated, when QM textbooks explain the general theory of angular momenta they are actually finding the irreps of ${\frak su}_{\mathbb{C}}(2)$. The result is that these representations are classified by a number $j\in \frac{1}{2}\mathbb{Z}_+$ which is either integer or half-integer.

That said, representations of the Lorentz algebra can be classified by two numbers $(A,B)\in \frac{1}{2}\mathbb{Z}_+\times \frac{1}{2}\mathbb{Z}_+$, each of which labels one of the ${\frak su}_{\mathbb{C}}(2)$ irreps. This is what is contained in your notation when you say that $(1,0)$ and $(0,1)$ are $2$-forms representations. I'll get back to the reason why they are $2$-forms in a moment.

On the other hand, given a vector space $V$ we can construct $(p,q)$-tensors over it. These are multilinear functionals of $p$ co-vectors and $q$ vectors: $$T:\underbrace{V^\ast\times \cdots V^\ast}_{\text{$p$ times}}\times \underbrace{V\times\cdots \times V}_{\text{$q$ times}}\to \mathbb{R}.$$

The $(0,p)$-tensors which are completely anti-symmetric are called $p$-forms and their space is denoted $\bigwedge^p V^\ast$. The important point here is that the various spaces $T_{(p,q)}V$ of $(p,q)$-tensors form representations of ${\rm GL}(n,\mathbb{R})$ where $n=\dim V$.

Now, the key observation is that ${\rm SO}(1,3)\subset {\rm GL}(4,\mathbb{R})$ is a subgroup. You may construct representations of ${\rm SO}(1,3)$ by restricting those of ${\rm GL}(4,\mathbb{R})$, i.e., you look at the group action just for the matrices in.${\rm SO}(1,3)$. This works for the Lorentz representations which are representations of ${\rm SO}(1,3)$ itself, i.e., not those which are just representations of the universal cover ${\rm SL}(2,\mathbb{C})$. In other words, this "structure group reduction" approach only works for the integer spin representations.

In particular, it is clear that the representations you get from ${\rm GL}(4,\mathbb{R})$ must also be described in terms of the first classification. Indeed, the vector representation is the $(\frac{1}{2},\frac{1}{2})$ representation while the $2$-form representation is the $(1,0)\oplus (0,1)$ representation.

So in summary, there are two notations in your question. The $(1,0)$ and $(0,1)$ representations are related to the $(A,B)$ classification of irreps of the Lorentz algebra in terms of angular momentum representations. The $(0,p)$ forms are related to the tensor representations of ${\rm GL}(4,\mathbb{R})$. For a subset of the Lorentz algebra representations these two can be identified. In particular, the relevant subset is the subset of integer spin representations.

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  1. On one hand, Carroll is talking about the order/rank of a tensor, i.e. the number of contravariant and covariant indices for its components, cf. e.g. this Phys.SE post.

  2. On the other hand, Wikipedia is talking about spin and/or dimensions of group representations for the Lorentz group in 3+1D, cf. e.g. this Phys.SE post.

We have the Minkowski metric to raise and lower indices, so by a $p$-form, we essentially mean the totally antisymmetric representation $$\bigwedge\!{}^pV,$$ where $$V~\cong~(\frac{1}{2},\frac{1}{2})$$ is the 4-vector representation.

Moreover in 3+1D it is enough to consider the cases $p=0,1,2$ because of Hodge duality.

For $p=0$, $$ \bigwedge\!{}^0V ~\cong~~(0,0). $$ For $p=1$, $$ \bigwedge\!{}^1V ~\cong~V~\cong~(\frac{1}{2},\frac{1}{2}). $$ For $p=2$, $$ \bigwedge\!{}^2V ~\cong~(1,0)~\oplus~ (0,1), $$ cf. e.g. eq. (3) in my Phys.SE answer here.

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  • $\begingroup$ So should I expect the $3$-form and $4$-form to be identified with four- and one-dimensional representations of the Lorentz group, particularly $\left(\tfrac{1}{2},\tfrac{1}{2}\right)$ and $\left(0,0\right)$? $\endgroup$ Commented Jan 5, 2023 at 1:51
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    $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Commented Jan 5, 2023 at 9:19

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