The orthogonality relations for representations of finite groups

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The orthogonality relations for representations of finite groups

August 30, 2009 by Qiaochu Yuan

In order to continue our discussion of symmetric functions it will be useful to have some group representation theory prerequisites, although I will use many of the results in the representation theory of the symmetric groups as black boxes. I had planned on using this post to discuss Frobenius reciprocity, but got so carried away with motivating it that this post now stands alone.

Today I’d like to discuss the representation theory of finite groups over $\mathbb{C}$ . As these are strong assumptions, the resulting theory is quite elegant, but I always found the proofs a little unmotivated, so I’m going to try to use the categorical perspective to fix that. Admittedly, I don’t have much experience with this kind of thing, so this post is for my own benefit as much as anyone else’s. The main focus of this post is motivating the orthogonality relations.

When talking about a subject as popular as representation theory, a math blogger runs the risk of repeating material that has been thoroughly exposited on other blogs. Here are some of the posts I will try to avoid repeating, at least not without citation: John Armstrong has written about the general category-theoretic picture, hilbertthm90 has some posts about the basic results of group representation theory, and Akhil Mathew over at Delta Epsilons has also written about group representations. (For the sake of completeness, there are also the higher-level representation-theoretic posts at the Secret Blogging Seminar and at Concrete Nonsense.)

Intertwining operators

Let $G$ be a finite group regarded as a one-object category and let $\text{FinVect}$ denote the category of finite vector spaces (here, over $\mathbb{C}$ ) and linear operators between them. The finite-dimensional representations of a group $G$ form a functor category $\text{Rep}(G)$ whose objects are functors $G \to \text{FinVect}$ . We will use capital uppercase letters such as $V$ to denote the underlying vector space of a representation and Greek letters such as $\rho$ to denote the function that sends an element of $G$ to a linear transformation $V \to V$ , and we will often name a representation by its underlying vector space. Concretely, a representation of $G$ on $V$ is a homomorphism $\rho : G \to \text{Hom}(V, V)$ (in $\text{FinVect}$ ; from here on in, $\text{Hom}$ should be interpreted as being in $\text{Rep}(G)$ ).

The morphisms in $\text{Rep}(G)$ are natural transformations of functors. Concretely, a morphism between two representations $\rho_1, \rho_2$ of the same group on vector spaces $V_1, V_2$ is a linear transformation $\phi : V_1 \to V_2$ such that $\phi \rho_1 = \rho_2 \phi$ . To avoid confusion regarding what “morphism” refers to we will call such a map an intertwining operator. Two representations are isomorphic if there is an invertible intertwining operator between them.

The category $\text{Rep}(G)$ is quite rigid. One can think of it as $\text{FinVect}$ “with extra structure” (just as, say, groups are sets with extra structure), and from that perspective $\text{Rep}(G)$ inherits the structure of an additive category, so it has a biproduct in the form of the direct sum of representations $V_1 \oplus V_2$ defined by setting $(\rho_1 \oplus \rho_2)(g) = \rho_1(g) \oplus \rho_2(g)$ . Additivity also implies that its Hom-sets are enriched over $\text{FinVect}$ . (All this means is that intertwining operators from $A$ to $B$ form a vector space.) The universal properties of the biproduct imply that for any three representations $A, B, C$ one has natural identifications

$\text{Hom}(A \oplus B, C) \simeq \text{Hom}(A, C) \oplus \text{Hom}(B, C) \\ \text{Hom}(A, B \oplus C) \simeq \text{Hom}(A, B) \oplus \text{Hom}(A, C)$ .

Anticipating later discussion we will write these as

$\dim \text{Hom}(A \oplus B, C) = \dim \text{Hom}(A, C) + \dim \text{Hom}(B, C) \\ \dim \text{Hom}(A, B \oplus C) = \dim \text{Hom}(A, B) + \dim \text{Hom}(A, C)$ .

Thus the $\text{Hom}$ functor is “bilinear.”

Concretely, let’s talk about how this works for $\text{FinVect}$ first. The direct sum $X \oplus Y$ is characterized by the fact that every vector in it is uniquely representable as a sum $x + y$ where $x \in X, y \in Y$ . (I’m abusing notation here, but it should be clear what I mean by this.) It is this uniqueness that is the key to its universal properties. If we ignore all the representation stuff and just consider linear operators, then:

Every linear operator $A \oplus B \to C$ satisfies $\phi(a + b) = \phi(a) + \phi(b), a \in A, b \in B$ , and $\phi(a)$ must be a linear operator on $A$ while $\phi(b)$ must be a linear operator on $B$ . Conversely, any pair of linear operators $A \to C, B \to C$ describes an operator $A \oplus B \to C$ . Hence the direct sum is a finite coproduct in $\text{Vect}$ .
Every linear operator $A \to B \oplus C$ has the property that the image $\phi(a)$ can be uniquely written as the sum of two functions $\phi_B(a) \in B, \phi_C(a) \in C$ , which must themselves be linear operators. Conversely, any pair of linear operators $A \to B, A \to C$ describes a linear operator $A \to B \oplus C$ . Hence the direct sum is a finite product in $\text{Vect}$ .

But it’s easy to verify that above arguments continue to hold if $\phi$ is required to be an intertwining operator rather than a mere linear operator; the operators $\phi(a), \phi(b), \phi_B, \phi_C$ must all in fact be intertwining.

A subobject in $\text{Rep}(G)$ is a subrepresentation, or a subspace invariant under the action of $G$ , and a representation is irreducible if it has no nontrivial subrepresentations. A second source of rigidity in $\text{Rep}(G)$ comes from Schur’s lemma, which can be stated (over $\mathbb{C}$ ) as follows: if $A, B$ are irreducible, then

$\dim \text{Hom}(A, B) = \begin{cases} 1 & \text{ if } A \simeq B \\ 0 & \text{ otherwise } \end{cases}$ .

Thus irreducible representations are “orthonormal.”

The proof is short enough to give here: the kernel of any intertwining operator $A \to B$ must be $G$ -invariant, hence it forms a subrepresentation of $A$ , which by irreducibility is either all of $A$ or $0$ . In the latter case $A, B$ are isomorphic, so one can regard an intertwining operator $\phi$ as an endomorphism of $A$ . Over $\mathbb{C}$ , this endomorphism has an eigenvector with some eigenvalue $c$ , and the nullspace of $\phi - cI$ is again $G$ -invariant, hence all of $A$ .

Combining Schur’s lemma and “bilinearity” gives the following result.

Proposition: Let $V_1, V_2, ...$ be non-isomorphic irreducible complex representations of a group $G$ , and let $A, B$ be semisimple, i.e. direct sums of irreducible representations $A = \bigoplus_{i =1}^{n} a_i V_i, B = \bigoplus_{i=1}^{n} b_i V_i$ where $a_i V_i$ indicates that the irreducible representation $V_i$ occurs as a summand $a_i$ times. Then

$\displaystyle \dim \text{Hom}(A, B) = \sum_{i=1}^{n} a_i b_i$ .

If the point hasn’t already been made, the $\text{Hom}$ functor in $\text{Rep}(G)$ has the remarkable property that it behaves like a categorified inner product. I believe some people call this structure a 2-Hilbert space, and I really like the circle of ideas here. As John Baez writes,

The inner product of a Hilbert space is a bilinear map

$\left< \cdot, \cdot \right> : \overline{x} \times x \to \mathbb{C}$

taking each pair of elements $v, w \in x$ to the inner product $\left< \cdot, \cdot \right>$ . Here $\overline{x}$ denotes the conjugate of the Hilbert space $x$ . Similarly, the hom functor in a category $C$ is a bifunctor

$\text{hom}(\cdot, \cdot) : C^{op} \times C \to \text{Set}$

taking each pair of objects $c, d \in C$ to the set $\text{hom}(c, d)$ of morphisms from $c$ to $d$ . This analogy clarifies the relation between category theory and quantum theory that is so important in topological quantum field theory. In quantum theory the inner product $\left< \cdot, \cdot \right>$ is a number representing the amplitude to pass from $v$ to $w$ , while in category theory $\text{hom}(c, d)$ is a set of morphisms passing from $c$ to $d$ .

This is a beautiful insight; I think intuition from physics is a fundamental tool in mathematics and I wish I knew more examples as nice as this.

Another important source of rigidity in $\text{Rep}(G)$ is Maschke’s theorem, which states that every representation of a finite group is semisimple. As a corollary, the previous results now completely describe the morphisms in $\text{Rep}(G)$ . I’m somewhat unsatisfied with the standard proof of Maschke’s theorem, so I’ll delay it until the end; it interrupts the flow of the post somewhat.

The orthogonality relations and character theory

The character theory of finite groups allows us to compute $\dim \text{Hom}(A, B)$ using surprisingly little information about the representations themselves. Associated to any representation $\rho$ is a function $\chi(g) = \text{tr } \rho(g)$ called the character of the representation. Given two characters $\chi_A, \chi_B$ of irreducible representations the orthogonality relations state that

$\displaystyle \left< \chi_A, \chi_B \right> = \frac{1}{|G|} \sum_{g \in G} \overline{ \chi_A(g) } \chi_B(g) = \begin{cases} 1 & \text{ if } A \simeq B \\ 0 & \text{ otherwise } \end{cases}$ .

By bilinearity it then follows that $\left< \chi_A, \chi_B \right> = \dim \text{Hom}(A, B)$ . In other words, in the case of $\text{Rep}(G)$ we can in fact associate a concrete inner product with our categorified inner product. The proofs of this result that I have seen do not mention explicitly the connection to $\dim \text{Hom}(A, B)$ and are annoyingly computational, so I would like to give a more conceptual (read: functorial) proof.

As a way of motivating it, here is a proof for permutation representations. Let $G$ act on two finite sets $X, Y$ and consider the corresponding representations on $\mathbb{C}^X, \mathbb{C}^Y$ , which have characters $\text{Fix}_X(g), \text{Fix}_Y(g)$ . By the orbit-counting lemma,

$\displaystyle \frac{1}{|G|} \sum_{g \in G} \text{Fix}_X(g) \text{Fix}_Y(g)$

is precisely the number of orbits in the product group action $g(x, y) = (gx, gy)$ on $X \times Y$ . Thus it remains to prove the following.

Proposition: The intertwining operators $\mathbb{C}^X \to \mathbb{C}^Y$ have a basis which is in one-to-one correspondence with orbits of the action of $G$ on $X \times Y$ .

Proof. Given any orbit $(x_1, y_1), ... (x_n, y_n)$ of $X \times Y$ there is an intertwining operator which sends $x_i$ to $y_i$ and sends everything else to zero. Conversely, suppose $\phi$ is an intertwining operator with matrix $\phi_{i, j}$ (where we pick the basis given by the elements of $X, Y$ ). Then the intertwining condition requires that, working through the definition of matrix multiplication,

$\phi_{y, gx} = \phi_{g^{-1} y, x}$

or equivalently that $\phi_{x, y}$ is constant on orbits of $X \times Y$ . The conclusion follows.

A proof for general representations

Let’s generalize the above proof. The generalization of the product group action is the (internal) tensor product $A \otimes B$ of representations, ~~but defining it properly requires recognizing a subtlety I usually ignore.~~ (Edit, 8/30/09: The text that used to be here has been moved to where it is more relevant; see Theo’s comment below.)

The tensor product $A \otimes B$ ~~of a right representation and a left representation~~ is a ~~left~~ representation defined on the tensor product of the underlying vector spaces with the action given by the linear extension of $g(a \otimes b) = (\rho_A(g) a) \otimes (\rho_B(g) b)$ . It satisfies the following universal property: any bilinear function $(\cdot, \cdot) : A \times B \to \mathbb{C}$ preserving the action of $G$ in the sense that $(a \rho_A(g), b) = (a, \rho_B(g) b)$ must factor through the tensor product, i.e. must come from an intertwining operator $A \otimes B \to \mathbb{C}$ (regarded as the trivial representation). As expected from a generalization of the product action, the tensor product has character $\chi_A \chi_B$ .

The generalization of the orbit-counting lemma is the following special case of the orthogonality relations: for any representation $V$ with character $\chi$ , $\frac{1}{|G|} \sum_{g \in G} \chi(g)$ counts the number of copies of the trivial representation in $V$ . Since adding a copy of the trivial representation increases the value of $\chi(g)$ by $1$ (in the permutation picture, every permutation gains a fixed point), it suffices to show that for a non-trivial irreducible representation one has

$\displaystyle \frac{1}{|G|} \sum_{g \in G} \chi(g) = 0$ .

One way to prove this is as follows: the image of the linear transformation $\sum_{g \in G} g$ is a $G$ -invariant subspace of dimension at most $1$ and hence (under the assumption of non-triviality) must be zero. I dislike this proof for the same reason that I dislike the usual proof of Maschke’s theorem, but I’ll let this one slide.

There is an additional step necessary to account for conjugate linearity. The definition of a representation we gave above is equivalent to giving a left action $G \times V \to V$ satisfying $\rho(h) ( \rho(g) v) = \rho(hg) v$ . We can also define a right action $V \times G \to V$ satisfying $(v \rho(g)) (\rho(h) = v \rho(gh)$ ; note that the order of multiplication is reversed. (Equivalently, a right action of $G$ is a left action of the opposite group $G^{op}$ .)

Any left representation $\rho$ of a group $G$ on a finite-dimensional vector space $V$ naturally (that is, functorially) defines a right dual representation $\overline{\rho}$ on the dual space $V^{*}$ , as follows: there is a unique representation $\overline{\rho}$ preserving the natural pairing $(\cdot, \cdot) : V^{*} \times V \to \mathbb{C}$ in the sense that $(v^{*}, v) = (v^{*} \overline{\rho(g)}, \rho(g) v)$ . In matrix form, if one thinks of the natural pairing as multiplying row and column vectors, $\overline{ \rho(g) } = \rho(g^{-1})$ , but acting on row vectors instead of column vectors; thus if we want to think of the dual representation as a left representation we should take the transpose of $\rho(g^{-1})$ , which recovers the usual order of multiplication. Because every element of a finite group has finite order, $\rho(g)$ is diagonalizable and has eigenvalues consisting of roots of unity for every $g$ , and roots of unity have the property that their inverses are their conjugates; in other words, the dual representation has character the conjugate of the character of $\rho$ . (This is essentially because representations of finite groups over $\mathbb{C}$ are isomorphic to unitary representations, which is most of the proof of Maschke’s theorem; more on this later.)

It now follows that for two representations $A, B$ the inner product $\left< \chi_A, \chi_B \right>$ is equal to the number of copies of the trivial representation in the representation $A^{*} \otimes B$ . Thus it remains to prove the following.

Proposition: $\text{Hom}(A, B)$ is isomorphic to the subspace of $A^{*} \otimes B$ consisting of the vectors fixed by all of $G$ .

Proof. Suspend the representation theory and again focus on the linear algebra. Any pure tensor $a^{*} \otimes b$ naturally defines a linear operator $A \to B$ defined by letting $a^{*}$ act on $A$ and multiplying the result by $b$ . Consequently one can show by induction or by picking a basis that every linear operator $A \to B$ is a linear combination of such pure tensors.

Now we reintroduce the representation theory. Generalizing the permutation proof, the condition that every element of $G$ fixes an element of $A^{*} \otimes B$ is precisely the condition that the corresponding linear operator $\phi$ satisfies $\phi \rho_A(g) = \rho_B(g) \phi$ ! (In other words, one way to motivate the definition of both the tensor and dual representations is that this should be true.) This completes the proof.

A corollary of the orthogonality relations is the surprising fact that a representation is determined by its character (up to isomorphism), since its decomposition into irreducible representations can be computed from the character of alone. Since characters are invariant under conjugation, another corollary is that the number of irreducible representations is exactly the number of conjugacy classes of $G$ .

Other perspectives

More general than the notion of representation of a group is a representation of an associative algebra, which can be defined as an algebra homomorphism $A \to \text{Hom}(V, V)$ (since $\text{Hom}(V, V)$ has the structure of an algebra with the multiplication given by composition). For the case of finite groups the relevant algebra is the group algebra $\mathbb{C}[G]$ , which consists of formal linear combinations of the elements of $G$ which multiply by the group operation. From this perspective, one way to think about the representation theory of finite groups is in terms of special properties enjoyed by group algebras.

Even more general than the notion of a representation of an associative algebra is that of a module for a ring. The module-theoretic perspective clarifies many of the above constructions, since the category $R-\text{Mod}$ of (left) modules over a fixed ring $R$ directly generalizes the category of vector spaces over a field. An intertwining operator of $G$ -representations is precisely a homomorphism of $\mathbb{C}[G]$ -modules, since the requirement that “scalar multiplication” be respected is precisely the requirement that the action of $G$ be respected. $R-\text{Mod}$ is an additive category for any ring $R$ , so one gets the “bilinearity” of the $\text{Hom}$ functor in a very general context. And the notion of irreducible representation (as well as the proof of Schur’s lemma) carries through to simple modules in general.

However, neither internal tensor products nor dual representations are defined for modules over general rings. Their existence is due to the fact that group algebras carry a lot of extra structure: they are in fact Hopf algebras. The comultiplication $\Delta(g) = (g, g)$ gives us the internal tensor product and the antipode gives us duals. The Wikipedia article explains this quite well.

The two major tools left to be explained are Maschke’s theorem and characters. Maschke’s theorem is essentially a compactness result. It generalizes straightforwardly to compact groups as follows: given a representation of a compact group $G$ on a Hilbert space $H$ with inner product $(\cdot, \cdot ) : H \times H \to \mathbb{C}$ one uses the Haar integral to average over $G$ , obtaining a new inner product which is $G$ -invariant. $G$ -invariant inner products have the property that the orthogonal complement of a $G$ -invariant subspace (that is, a subrepresentation) is also $G$ -invariant, so it follows by induction that every finite-dimensional representation is semisimple, since any finite-dimensional vector space can be turned into a Hilbert space. (The standard results are much stronger than this, but I don’t know much about them.) As a corollary, every compact subgroup of $GL_n(\mathbb{C})$ is conjugate to a subgroup of the unitary group $U_n(\mathbb{C})$ ; in other words, the unitary group is an essentially unique maximal compact subgroup of $GL_n(\mathbb{C})$ .

Characters, on the other hand, are still mysterious to me. There is a functorial definition of the trace based on the identification of $\text{Hom}(V, V)$ with $V^{*} \otimes V$ , then extending the natural bilinear pairing $V^{*} \times V \to \mathbb{C}$ , but I can’t come up with a particularly good reason why this should accurately reflect the structure of a representation. If anyone has any comments on this, they would be much appreciated.

Posted in math.GR, math.RT | Tagged universal properties | 12 Comments

12 Responses

on March 21, 2021 at 3:01 pm | Reply edeany

You mention being unsatisfied with your proof (“I dislike this proof for the same reason that I dislike the usual proof of Maschke’s theorem”). I sense there is a different approach which would stem from observing that the character of an irreducible representation preserves products. Product preserving maps in the Hopf algebra $[G, \mathbb{C}]$ are always orthogonal for $G$ finite.
on September 3, 2009 at 10:31 am | Reply Josh

Akhil: although Qiaochu’s post gives a very nice exposition of some of classical representation theory, I suspect it is only very nice for those already well-versed in either representation theory or category theory. Although I agree that this is a much nicer presentation than in several books already mentioned in the comments, I do not see how this category-theoretic viewpoint answers the question “Why is this interesting?”
- on September 3, 2009 at 10:51 am | Reply Qiaochu Yuan
  
  Point taken. This post wasn’t intended to be a beginner’s introduction; I very much had in mind an audience who had seen the basic results at least once, and I wasn’t trying to motivate interest in the subject. Hopefully my own motivation regarding the representation theory of the symmetric groups will become clear in a few posts.
on August 30, 2009 at 4:12 pm | Reply Theo

Nice post! A few comments:

I’m bothered by the second paragraph under “A proof for general representations”. The problem is the left/right thing. The category of _left_ G-reps is monoidal, where on objects the monoidal structure is \otimes_\CC and the action of g on V\otimes W is \rho_V(g) \otimes \rho_W(g), where \rho_V,\rho_W are the actions on V,W. Similarly, the category of _right_ G-reps is monoidal. In either case, the one-dimensional rep \CC with the trivial action is the monoidal unit.

You propose an a tensor product that multiplies a right G-rep with a left one. This will not define a representation of G (unless G is abelian, whence “right” and “left” are the same). What you can do is this: if R is any ring (perhaps \CC[G], the group algebra), and V is a right R-module and W is a left R-module, then the product V \otimes_R W is defined. However, it is not a module over R at all — the R action has canceled out. If you want an R-linear monoidal category, the easiest thing to do is to take the category of R-R bimodules, i.e. spaces with both a left R action and a right R action. (If R is commutative, these can be required to be the same, or not, depending on your goals.) The morphisms are required to commute with both actions, and the two actions are requires to commute. This category is monoidal with \otimes_R as the monoidal structure, and R with its canonical left- and right- actions on itself is the monoidal unit.

Ok, so it looks like you wanted that right representation in order to say that \Hom(V,W) = V^* \otimes W. The trick is this. If V is a left module over R, then V^* is a right R-module with the transpose action. How do you turn a right R-module into a left R-module? In general you cannot — doing so functorially requires R to have an antiautomorphism, i.e. an additive map s: R\to R satisfying s(ab) = s(b) s(a).

Fortunately, in the category of representations of a group, you have one of these: the map G \to G sending g \mapsto g^{-1} extends to an antiautomorphism of the group algebra \CC[G]. So this lets you define on V^* a _left_ G-action, namely g \mapsto \rho(g^{-1})^*, where \rho(g) is the action of g on V.

If you’re doing Hopf algebras, the map s is called the “antipode”. There is one more condition on the antipode in a Hopf algebra: it is required to be an antiautomorphism of the comultiplication as well as of the multiplication. This assures that (V\otimes W)^* = W^* \otimes V^*, which it should be. In the group algebra, the comultiplication is almost trivial, and so this condition is free.

Anyhoo, so all of this defines on \Hom_\CC(V,W) the structure of a left G-module, if V and W are both left G-modules. And then the action is that g acts on a linear map \phi by taking it to \rho(g) \circ \phi \circ \rho(g^{-1}), so you’re exactly correct that the G-intertwiners \Hom_G(V,W) are precisely the fixed points of the G-action \Hom_\CC(V,W).

A good interpretation of this is as follows. \CC with the trivial action is the monoidal unit in G-rep, and \Hom_G(\CC,V) is the vector space (not naturally a G-rep) of fixed points of the G-action on V. If we replace the word “G-rep” by, say, “sheaf”, then \Hom_G(\CC,V) is “the space of global sections of V”; this is a useful notion in any closed monoidal category, and the discussion above shows that G-rep is closed. So we’re saying that the global sections of the enriched \Hom(V,W) comprise precisely the set of morphisms \Hom(V,W). This is true in an arbitrary closed monoidal category.
- on August 30, 2009 at 6:48 pm | Reply Qiaochu Yuan
  
  Thanks for the comment! That’s the part I was wondering about – the left/right convention was taken from the Wikipedia article about tensor products of modules, but it seemed out of place here after I thought about it. I’ll correct that part.
on August 30, 2009 at 2:59 pm | Reply John Armstrong

I’ll have to look this over later, but it seems at first glance that it might be useful as a reference when I finally get around to more finite-group representation theory.
- on August 30, 2009 at 3:11 pm | Reply Qiaochu Yuan
  
  I wouldn’t trust my definitions to be exactly right, since I’m not working off of a reference here; in particular I may be somewhat confused about the tensor product.
on August 30, 2009 at 2:46 pm | Reply Akhil Mathew

Nice post.

Most people who write about representation theory (including, unfortunately, myself in what was supposed to be a quick summary though) tend not to emphasize the categorical nature of all this. As a different (and more trivial) example from what you mentioned, if $F: \mathrm{FinVect} \to \mathrm{FinVect}$ is a (say covariant additive) functor, it induces a functor on $Rep(G) \to Rep(G)$ just by composition. Similarly for contravariant functors and bifunctors. Thus the facts that tensor products and hom-sets of representations are representations all become immediate rather than requiring separate ad-hoc arguments.

I think this tends to be a problem in textbooks, which often just pressent the material with minimal motivation and without answering the question “Why is this interesting?” especially when the material becomes so ad hoc.
- on August 30, 2009 at 3:09 pm | Reply Qiaochu Yuan
  
  I agree. For example, Artin’s proof of the orthogonality relations involves a lot of juggling around terms in sums, which I’ve never found a satisfying way to convince myself of anything. To my mind, the categorical perspective is just a generalization of the idea that binomial coefficient identities, for example, should be proven bijectively.
  - on August 30, 2009 at 3:17 pm | Reply Akhil Mathew
    
    Ditto for Serre, which means I forget the proof quickly. Fulton and Harris reduce to the case where one of the irreducible modules is 1, which I think is an improvement.
    
    I think Lang’s Algebra proves them a bit more functorially though.
- on August 30, 2009 at 8:31 pm | Reply Steven Sam
  
  I have a vector space V and I look at the functor which is tensor by V, the representation I get on $V \otimes W$ will be trivial on the V and whatever it was on W. How do you get an arbitrary action on V to carry over via a tensor product functor?
  - on August 31, 2009 at 3:47 am | Reply Akhil Mathew
    
    I think you have to consider the bifunctor $\mathrm{Vect} \times \mathrm{Vect} \to \mathrm{Vect}$ given by $V, W \to V \otimes W$ (and similarly for $\hom$ ).

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Annoying Precision

"A good stock of examples, as large as possible, is indispensable for a thorough understanding of any concept, and when I want to learn something new, I make it my first job to build one." – Paul Halmos

The orthogonality relations for representations of finite groups

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The orthogonality relations for representations of finite groups

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