# Abstract Nonsense

## Matrix Functions Form an (almost) Orthonormal Basis (Pt. II)

Point of post: This is a continuation of this post.

Matrix Entry Functions Span the Group Algebra

We now show that $\Lambda=\text{span }\mathcal{D}=\mathcal{A}(G)$. To do this we prove a result similar to that of the Stone-Weierstrass theorem for finite sets. Namely:

Theorem: Let $X$ be some finite set and $\mathscr{A}$ a subalgebra of $\mathbb{C}^X$ which separates points, then $\mathscr{A}=\mathbb{C}^X$. (here the algebra structure is the usual pointwise scalar multiplication, pointwise addition, and pointwise product of functions).

Proof: Note that $\left\{\delta_x\right\}_{x\in X}$ is a basis for $X$ where $\delta_x(y)=\delta_{x,y}$. Note though, for each $y\in X$ we are guaranteed some $f_{x,y}\in\mathscr{A}$ such that $f(x)=1$ and $f(y)=0$. It’s evident though that

$\displaystyle \delta_x=\prod_{y\ne x}f_{x,y}$

and so in particular since $\mathscr{A}$ is a subalgebra we have that $\delta_x\in\mathscr{A}$ for every $x\in X$ and since $\mathscr{A}$ is, in particular, a subspace we have that $\mathscr{A}\supseteq\text{span}\{\delta_x\}_{x\in X}=\mathbb{C}^X$. The conclusion follows. $\blacksquare$

From this we pretty easily derive the desired result. Namely:

Theorem: $\Lambda=\mathcal{A}(G)$.

Proof: We’ve proven in previous posts that $\Lambda$ is a subalgebra of $\mathbb{C}^G=\mathcal{A}(G)$ (this equality is as a set, since the group algebra has a different multiplicative structure than pointwise multiplication) which separates points, and thus the conclusion follows by the previous theorem. $\blacksquare$

Corollary: $\mathcal{D}$ is a basis for $\mathcal{A}(G)$ and thus in particular $\#\left(\widehat{G}\right)<\infty$ and

$\displaystyle \sum_{\alpha\in\widehat{G}}d_\alpha^2=\#\left(\mathcal{D}\right)=\dim\mathcal{A}(G)=|G|$

Note though that since we also have the trivial irrep $\tau:G\to\mathbb{C}:g\mapsto 1$ that the above implies that, for example, any non-trivial irrep $\rho$ has the quality that $\deg\rho\leqslant \sqrt{|G|-1}$. We shall use this to derive many, many more interesting result as the theory progesses.

References:

1.Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

2. Serre, Jean Pierre. Linear Representations of Finite Groups. New York: Springer-Verlag, 1977. Print

February 23, 2011 -

## 9 Comments »

1. […] Let . Since in particular and as was already proven. Thus, there exists constants for every and such […]

Pingback by Representation Theory Irreducible Characters (Pt. II) « Abstract Nonsense | February 25, 2011 | Reply

2. […] if we can prove that is unitary then it will be injective, but we know that the domain and codomain are of finite and equal dimension and so is an injective if and only […]

Pingback by Representation Theory: Decomposing the Group Algebra Into the Direct Sum of Matrix Algebras « Abstract Nonsense | April 6, 2011 | Reply

3. […] Proof: We know that it suffices to prove that every irrep of is of degree one. So, we know that […]

Pingback by Groups of Order pq (pt. I) « Abstract Nonsense | April 19, 2011 | Reply

4. […] Proof: Since is odd we know that every non-trivial irrep of is complex. Thus, for every in one has that and . But, since is odd and we know that we may conclude that for some . Thus, since we know that the cardinality of is (where ) we enumerate the elements of as . Thus, we have by prior theorem that […]

Pingback by Representation Theory: The Number of Conjugacy Classes of a Finite Group of Odd Order is Equivalent to The Order of The Group Modulo Sixten « Abstract Nonsense | April 23, 2011 | Reply

5. […] We know that (we know that is equal to the conjugacy class of since it is the only trivial conjugacy […]

Pingback by University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) (August-2003) « Abstract Nonsense | May 1, 2011 | Reply

6. […] tableaux on . In this post we prove a result which is not only integral in proving this fact but is consistent with this hypothesis, namely that the sum over all -frames with is […]

Pingback by The Fundamental Result for Tableaux Combinatorics « Abstract Nonsense | May 12, 2011 | Reply

7. Hey Alex,

Nice blog!
I’m starting one here:
http://dmitrigekhtman.wordpress.com/
but I’m kind of inexperienced.

Would you mind sending me the code for this post?
I’d like to figure out how to use WordPress’s LaTeX features more effectively.
My e-mail address is dmitrgekh@gmail.com.

Thanks,
Dmitri

Comment by Dmitri Gekhtman | July 4, 2011 | Reply

• Hello Dmitri!

Thank you very much for your support! I’m glad to see that you will be a new memeber to the fantastic wordpress.com blog community! I am having a bit of e-mail trouble at the moment, so I hope copy and pasting the code at the bottom of this response will suffice.

Anyways, tell me a little bit about yourself. What’s your mathematical background, interests, etc.?

Best,
Alex

Point of post: This is a continuation of this post.

Matrix Entry Functions Span the Group Algebra

We now show that $\Lambda=\text{span }\mathcal{D}=\mathcal{A}(G)$. To do this we prove a result similar to that of the Stone-Weierstrass theorem for finite sets. Namely:

Theorem: Let $X$ be some finite set and $\mathscr{A}$ a subalgebra of $\mathbb{C}^X$ which separates points, then $\mathscr{A}=\mathbb{C}^X$. (here the algebra structure is the usual pointwise scalar multiplication, pointwise addition, and pointwise product of functions).

Proof: Note that $\left\{\delta_x\right\}_{x\in X}$ is a basis for $X$ where $\delta_x(y)=\delta_{x,y}$. Note though, for each $y\in X$ we are guaranteed some $f_{x,y}\in\mathscr{A}$ such that $f(x)=1$ and $f(y)=0$. It’s evident though that

$\displaystyle \delta_x=\prod_{y\ne x}f_{x,y}$

and so in particular since $\mathscr{A}$ is a subalgebra we have that $\delta_x\in\mathscr{A}$ for every $x\in X$ and since $\mathscr{A}$ is, in particular, a subspace we have that $\mathscr{A}\supseteq\text{span}\{\delta_x\}_{x\in X}=\mathbb{C}^X$. The conclusion follows. $\blacksquare$

From this we pretty easily derive the desired result. Namely:

Theorem: $\Lambda=\mathcal{A}(G)$.

Proof: We’ve proven in previous posts that $\Lambda$ is a subalgebra of $\mathbb{C}^G=\mathcal{A}(G)$ (this equality is as a set, since the group algebra has a different multiplicative structure than pointwise multiplication) which separates points, and thus the conclusion follows by the previous theorem. $\blacksquare$

Corollary: $\mathcal{D}$ is a basis for $\mathcal{A}(G)$ and thus in particular $\#\left(\widehat{G}\right)<\infty$ and

$\displaystyle \sum_{\alpha\in\widehat{G}}d_\alpha^2=\#\left(\mathcal{D}\right)=\dim\mathcal{A}(G)=|G|$

Note though that since we also have the trivial irrep $\tau:G\to\mathbb{C}:g\mapsto 1$ that the above implies that, for example, any non-trivial irrep $\rho$ has the quality that $\deg\rho\leqslant \sqrt{|G|-1}$. We shall use this to derive many, many more interesting result as the theory progesses.

References:

1.Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

2. Serre, Jean Pierre. Linear Representations of Finite Groups. New York: Springer-Verlag, 1977. Print

Comment by Alex Youcis | July 7, 2011 | Reply

8. Hmm, that didn’t work so well, did it? Haha. I’ll erase the  stuff. I assume you know you have to encase your LaTeX in $latex(space)(LaTeX)$ where obviously (space) means you need to put a space and (LaTeX) is whatever you want. I hope this is what you wanted. If you were more concerned with formatting (spaces, bolding, italicizing, etc.) or if you have any more questions, just let me know! Anyways:

Point of post: This is a continuation of this post.

Matrix Entry Functions Span the Group Algebra

We now show that \Lambda=\text{span }\mathcal{D}=\mathcal{A}(G). To do this we prove a result similar to that of the Stone-Weierstrass theorem for finite sets. Namely:

Theorem: Let X be some finite set and \mathscr{A} a subalgebra of \mathbb{C}^X which separates points, then \mathscr{A}=\mathbb{C}^X. (here the algebra structure is the usual pointwise scalar multiplication, pointwise addition, and pointwise product of functions).

Proof: Note that \left\{\delta_x\right\}_{x\in X} is a basis for X where \delta_x(y)=\delta_{x,y}. Note though, for each y\in X we are guaranteed some f_{x,y}\in\mathscr{A} such that f(x)=1 and f(y)=0. It’s evident though that

\text{ }

\displaystyle \delta_x=\prod_{y\ne x}f_{x,y}

\text{ }

and so in particular since \mathscr{A} is a subalgebra we have that \delta_x\in\mathscr{A} for every x\in X and since \mathscr{A} is, in particular, a subspace we have that \mathscr{A}\supseteq\text{span}\{\delta_x\}_{x\in X}=\mathbb{C}^X. The conclusion follows. \blacksquare

\text{ }

From this we pretty easily derive the desired result. Namely:

Theorem: \Lambda=\mathcal{A}(G).

Proof: We’ve proven in previous posts that \Lambda is a subalgebra of \mathbb{C}^G=\mathcal{A}(G) (this equality is as a set, since the group algebra has a different multiplicative structure than pointwise multiplication) which separates points, and thus the conclusion follows by the previous theorem. \blacksquare

Corollary: \mathcal{D} is a basis for \mathcal{A}(G) and thus in particular \#\left(\widehat{G}\right)<\infty and

\displaystyle \sum_{\alpha\in\widehat{G}}d_\alpha^2=\#\left(\mathcal{D}\right)=\dim\mathcal{A}(G)=|G|

Note though that since we also have the trivial irrep \tau:G\to\mathbb{C}:g\mapsto 1 that the above implies that, for example, any non-trivial irrep \rho has the quality that \deg\rho\leqslant \sqrt{|G|-1}. We shall use this to derive many, many more interesting result as the theory progesses.

References:

1.Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

2. Serre, Jean Pierre. Linear Representations of Finite Groups. New York: Springer-Verlag, 1977. Print

Comment by Alex Youcis | July 7, 2011 | Reply