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## Irreducible representations of a product of two groups

Let $V$ be an irreducible finite dimensional complex representation of the product of groups $G\times H$ . Is it necessarily isomorphic to a tensor product of irreducible representation of $G$ and $H$ ? If not what is a counter-example, and under what extra assumptions this is known to be true?

Remark. I think for continuous representations of compact groups this is true.

- gr.group-theory
- rt.representation-theory

- 2 $\begingroup$ For algebraically irreducible representations this is true and can be found in many sources like Curtis and Reiner. For any finite dimensional algebras the irreducible representations of the tensor product are the tensor products of irreducibles. You can replace the group algebra by the tensor product of the images of CG and CH to reduce to the finite dimensional case. $\endgroup$ – Benjamin Steinberg Dec 11, 2022 at 12:38
- 2 $\begingroup$ There is some discussion in comments to this question mathoverflow.net/questions/397592/… which might be useful (I see that @BenjaminSteinberg made many of the same points over there) $\endgroup$ – Yemon Choi Dec 11, 2022 at 15:41

## 2 Answers 2

If your groups are finite, then Andy’s answer is perfectly fine. If you want to do topological groups you must be slightly (but not much) more careful. First since we are dealing with finite dimensional representations, algebraic and topological reducibility are the same.

It is enough to prove that your irreducible representation decomposes as a tensor product in the case of discrete groups. For if it is isomorphic to one of the form $U\otimes V$ , then since the linear isomorphism is continuous, it is enough to observe that restricting the original representation to $G$ gives $\dim V$ copies of $U$ and restricting to $H$ gives $\dim U$ copies of $V$ and so $U$ and $V$ must give continuous representations of $G$ , $H$ .

So we may assume that $G$ , $H$ are discrete and we are dealing with no topology. Then replacing $G$ , $H$ with their group algebras and $\mathbb C$ by algebras over an algebraically closed field, it suffices to show that if $A$ , $B$ are $K$ -algebras with $K$ algebraically closed, then any finite dimensional irreducible representation of $A\otimes B$ is equivalent to a tensor product of irreducible representations of $A$ and $B$ .

But the image of $A$ and $B$ under any finite dimensional representation is a finite dimensional algebra and so without loss of generality we may assume that $A$ , $B$ are finite dimensional.

The crux of the matter, which is basically Andy’s proof in different words, is the special case that $A,B$ are matrix algebras over $K$ . Then $M_n(K)\otimes M_m(K)\cong M_{nm}(K)$ and the isomorphism intertwines the actions on $K^n\otimes K^m\cong K^{nm}$ . Thus in the case $A$ , $B$ are matrix algebras then the unique $A\otimes B$ irreducible rep is the tensor product of the unique irreducible $A$ and $B$ reps. Since arbitrary semisimple algebras over an algebraically closed field are finite direct products of matrix algebras and tensor product distributes over direct product, this handles the case $A$ , $B$ are semisimple and also shows that $A\otimes B$ is semisimple in this case.

$\DeclareMathOperator\rad{rad}$ If $A$ , $B$ are general finite dimensional $K$ -algebras with $K$ -algebraically closed, then $A/{\rad(A)}\otimes B/{\rad(B)}$ is semisimple by the above. Moreover, the kernel of the natural map $A\otimes B\to A/{\rad(A)}\otimes B/{\rad(B)}$ is easily checked to be $A\otimes \rad(B)+\rad(A)\otimes B$ , which is a nilpotent ideal. Thus $(A\otimes B)/{\rad(A\otimes B)}\cong A/{\rad(A)}\otimes B/{\rad(B)}$ and so the desired result that irreducible reps are tensor products follows since $C$ and $C/{\rad(C)}$ have the same irreducible reps for any $K$ -algebra $C$ .

Simpler proof Here is an argument along the line of Andy's that works for arbitrary groups or even algebras that avoids the radical and just uses Burnside's theorem that a finite dimensional representation of a $K$ -algebra over an algebraically closed field $K$ is irreducible if and only if it is surjective.

Let $A,B$ be $K$ -algebras (not necessarily finite dimensional) with $K$ -algebraically closed (they could be group algebras) and let $W$ be a finite dimensional irreducible $A\otimes B$ -module. Let $U$ be an irreducible $A$ -subrepresentation of $W$ . Then the sum $S$ of all irreducible $A$ -submodules of $W$ isomorphic to $U$ is invariant under any $A$ -enomorphism of $W$ and hence under $B$ as the $B$ -action commutes with $A$ . Hence $S$ is $A\otimes B$ -invariant and so $S=W$ by irreducibility. Thus $W\cong U^m$ for some $m$ as an $A$ -module (this is a standard argument). Thus, up to isomorphism, we may assume that $W=U\otimes V$ with $V$ a vector space of dimension $m$ and where $A$ acts via matrices of the form $\rho(a)\otimes 1$ where $\rho$ is the irreducible representation associated to $U$ . By Burnside, $\rho$ is onto $End_k(U)$ . But since $End_K(U\otimes V)= End_k(U)\otimes End_k(V)$ , it follows that the centralizer of $End_K(U)\otimes 1$ in $End_K(W)$ is $1\otimes End_k(V)$ . Since the action of $B$ commutes with that of $A$ , as a representation of $B$ , we have $W$ is of the form $b\mapsto 1\otimes \psi(b)$ for some representation $\psi$ of $B$ on $V$ . Clearly, if $V$ is not irreducible, then neither is $U\otimes V$ and so we are done.

Most books on the representation theory of finite groups prove this using character theory. This is an efficient way of proving it, but I think it doesn’t give much insight into why it is true (and what other settings it generalizes to). I wrote up a more direct proof here .

- 2 $\begingroup$ The OP doesn't say that the group here is finite. In fact he may be looking at topological irreducibility which for compact groups is the same and I guess this is true in finite dimensions for all topological groups since all subspaces are closed but maybe one has to check continuity. Anyway for reps of discrete infinite groups you need to argue differently because you don't get for free complete reducibility but it isn’t bad. Nonetheless is true as I argue in the comments. The argument for finite dimensional algebras is basically the same as your note after you factor out the radicals. $\endgroup$ – Benjamin Steinberg Dec 11, 2022 at 14:58
- 2 $\begingroup$ Actually I think your proof is fine for infinite groups with a tiny modification. You can argue that the restriction of an irreducible to a direct factor (or any normal subgroup) is completely reducible by noting the sum of all the translates of an irreducible subrep of the normal subgroup is the whole representation by irreduciblity and then proceed as in your proof. I'll keep mine since it works for monoids or for any algebras $\endgroup$ – Benjamin Steinberg Dec 11, 2022 at 15:41

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## Irreducible Representation

1. The dimensionality theorem :

3. Orthogonality of different representations

4. In a given representation, reducible or irreducible, the group characters of all matrices belonging to operations in the same class are identical (but differ from those in other representations).

6. A one-dimensional representation with all 1s (totally symmetric) will always exist for any group .

7. A one-dimensional representation for a group with elements expressed as matrices can be found by taking the group characters of the matrices .

Irreducible representations can be indicated using Mulliken symbols .

Portions of this entry contributed by Todd Rowland

## Explore with Wolfram|Alpha

More things to try:

- 30-sided polyhedron
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## Cite this as:

Rowland, Todd and Weisstein, Eric W. "Irreducible Representation." From MathWorld --A Wolfram Web Resource. https://mathworld.wolfram.com/IrreducibleRepresentation.html

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If V is a finite dimensional irreducible representation of G G, then it is well known that V V is a tensor product of Vi V i, i = 1, 2 i = 1, 2 and each Vi V i is an irreducible representation of Gi G i. The question I have is when V V is given, is there a canonical way to construct Vi V i from V V? rt.representation-theory Share Cite

1 I would like to know (how to compute) the irreducible representations resulting from the decomposition of tensor-product representations of product groups. For example: Consider the group SO6 × SO6 S O 6 × S O 6.

As the irreducible rep. of S3 S 3 are of dimension 1 or 2 only, this means that the resulting dimension of the tensor product is of dimension 2k 2 k for some k k.

For any finite dimensional algebras the irreducible representations of the tensor product are the tensor products of irreducibles. You can replace the group algebra by the tensor product of the images of CG and CH to reduce to the finite dimensional case. - Benjamin Steinberg Dec 11, 2022 at 12:38 2

In mathematics, specifically in the representation theory of groups and algebras, an irreducible representation or irrep of an algebraic structure is a nonzero representation that has no proper nontrivial subrepresentation , with closed under the action of .

IRREDUCIBLE REPRESENTATIONS OF THE SYMMETRIC GROUP REDMOND MCNAMARA Abstract. We construct the Specht modules and prove that they completely characterize the irreducible representations of the symmetric group.

The tensor product of two irreducible representations of a group or Lie algebra is usually not irreducible. It is therefore of interest to attempt to decompose into irreducible pieces. This decomposition problem is known as the Clebsch-Gordan problem. The SU (2) case

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information about the representations of the group. The first conjugacy class, C 1, is usually taken to be the one-element conjugacy class Thus, the elements Of the first column consist of the values Zi(e) = m, the degree of the ith irreducible representation. The character is usually taken to be the trivial representation, so that the entries of

Irreducible representations of product of real reductive groups Dmitry Gourevitch, Alexander Kemarsky Abstract. Let G 1;G 2 be real reductive groups and (ˇ;V) be a smooth admissible representation of G 1 G 2. We prove that (ˇ;V) is irreducible if and only if it is the completed tensor product of (ˇ i;V i), i = 1;2, where (ˇ i;V

tensor products of irreducible representations of Gand H. Because our proof avoids ... This is an irreducible representation; however, despite the fact that Z/6 ∼=Z/3 ×Z/2, it ... By assumption, V and W are simple modules over the group rings k[G] and k[H], respectively.

• The character of a tensor-product representation D ˆ⌦ D0 is the product of the respective ... which means that the characters of inequivalent irreducible representations of a given group G are orthogonal — and in fact orthonormal — for the scalar product (III.10). For every element g 2G, let us denote [g] the conjugacy class of g in ...

The representation theory of groups is a part of mathematics which examines how groups act on given structures. Here the focus is in particular on operations of groups on vector spaces. Nevertheless, groups acting on other groups or on sets are also considered. For more details, please refer to the section on permutation representations .

The irreducible representations of a point group satisfy a number of orthogonality relationships: 1. If corresponding matrix elements in all of the matrix representatives of an irreducible representation are squared and added together, the result is equal to the order of the group divided by the dimensionality of the irreducible representation ...

into irreducible polynomials, each of whose multiplicity equals its degree.) The history of group representation theory is explained in a book by Curtis, Pioneers of representation theory. This theory appears all over the place, even before its origin in 1896: In its origin, group theory appears as symmetries. This dates at least to Felix Klein's

The product is actually the irreducible representation given by \(A_ {2}\) ! As it turns out, the direct product will always yield a set of characters that is either an irreducible representation of the group, or can be expressed as a sum of irreducible representations. This suggests that a multiplication table can be constructed.

The direct product of two irreducible representations give either a reducible or irreducible representation of the same group. The last column in the character table shows the direct product between any two linear vectors. Direct products can also be taken between any number of irreducible representations: \[\Gamma_a \Gamma_b \Gamma_c=\Gamma ...

For our purposes a representation of a group Gis a collection R of linear operators on a complex vector space, together with a homomorphism φfrom Gto Rin which group multiplication is mapped to the product of operators, and group inverse to the inverse of the operator. Hence φ maps the identity eof Gto the identity operator Ion the vector space.

Only the direct product of a representation with itself is or contains the totally symmetric representation. 7.6: Characters of Irreducible Representations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Jerry LaRue & Claire Vallance. the character of a group representation is a function on the group that ...

No headers. Similarity transformations yield irreducible representations, Γ i, which lead to the useful tool in group theory - the character table.The general strategy for determining Γ i is as follows: A, B and C are matrix representations of symmetry operations of an arbitrary basis set (i.e., elements on which symmetry operations are performed). ). There is some similarity transform ...

Often ths process is simple, especially when one or both of the irreducible representations are non-degenerate (in most cases A or B ). This process is embodied in an equation. Groups where there are subscripts 1, 2 and 3 (eg D2 and D2h) - here 1x2=3, 2x3=1 and 1x3=2. For point groups where the principal axis is C2 or C4 (eg C4 and D2d ), E × ...