## Algebra parameters [on hold]

I got 4 equations, where $$x,y,z,u$$ are variables , and $$a,b$$ parameters. So the solution from the book is like this: $$x+(a+1)y-(a+1)z-au=1$$ $$ax+(a+1)y+az-2u=2$$ $$ax+(a+1)y-2z+au=b$$ $$(a-1)x+3(a+1)z-4u=3-b$$ So when I do all transformations it gets this form:

(a+1)y+x-(a+1)z-au=1

(a-1)x+(2a+1)z+(a-2)u=1

(-a-2)z+(a+2)u=b-2

Now, next steps are these:

1) If a#-1 , a#1 , a#-2
Equation is indefinite

2)If a=-1 Equation is indefinite

3)If a=1 Equation is indefinite

4)If a=-2

Equation can be either indefinite or impossible

My question is how did we got that first part where a can’t be 1, -1 or -2. Why it can be 2 or 0 or some other number?

## Dimensions of Lie algebra powers of irreducible representations

Consider the following plethysms. For semisimple Lie algebras $$L$$, if $$A$$ is the adjoint irrep, as far as I know, $$d_1=|A^{\bigotimes 1}|=0$$ (no lollipops), $$d_2=|A^{\bigotimes 2}|=1$$ (Schur’s Lemma), $$d_3=|A^{\bigotimes 3}|=1$$ (structure constant tensor, 6j symbols).
And if $$A$$ is the adjoint of a non-semisimple $$L$$? I also read something like that the Killing form $$K$$ can be brought to $$\mathrm{diag}(\{0,1\})$$ in the right basis, implying $$K^2=K$$ and thus $$d_2=2$$ (unless $$L$$ is nilpotent, $$K=0$$ and thus $$d_2=1$$). But I’m confused – rescale the basis and the Killing form might have more than two different eigenvalues for more than 4 generators? (Does it?)
Assuming it miraculously hasn’t, this means $$d_3\leq 8$$ (one or no Killing form pasted on the structure constant tensor). I already once jumped to conclusions ๐ and thought $$d_3=1$$, but I found a $$L$$ with $$d_3=4$$. So, can you name a $$L$$ with $$d_3=8$$?.
(In graphic form the answers to my question seem obvious ๐

## Division algebra over rationals of dimension 9

I want to understand about existence of some non-commutative division algebras over $$\mathbb{Q}$$ of dimension $$9$$.

Q. Does there exist a division algebra $$D$$ such that

• $$D$$ is non-commutative;

• $$D$$ is of dimension $$9$$ over $$\mathbb{Q}$$ and $$Z(D)=\mathbb{Q}$$;

• $$K:=\mathbb{Q}(2^{1/3})$$ is a maximal sub-field of $$D$$?

My way towards solution: if $$D$$ is such algebra, then consider $$x\in D$$ outside $$K:=\mathbb{Q}(2^{1/3})$$. If conjugation by $$x$$ leaves $$K$$ invariant then it induces an automorphism of $$K$$ which fixed $$\mathbb{Q}$$; the only possibility of this is trivial automorphism, which means $$x$$ centralizes $$K$$, contradiction.

In fact, we can show that $$D=K\oplus xK \oplus x^2K$$ as a vector space.

Further $$x^3$$ centralize all generators of $$D$$, so $$x^3\in\mathbb{Q}\setminus \{0\}$$. Next, how should I proceed to determine structure of $$D$$?

I never studied division algebras other than quaternions and fields. I do not know how this question will be, but I was trying to see whether after $$2^2$$, can we get non-commutative division algebra whose dimension over its center is $$3^2$$? So a simple case I thought is through above questions, I was unable to complete the solution of existence.

## Division algebra over rationals of dimension 9

I want to understand about existence of some non-commutative division algebras over $$\mathbb{Q}$$ of dimension $$9$$.

Q. Does there exist a division algebra $$D$$ such that

• $$D$$ is non-commutative;

• $$D$$ is of dimension $$9$$ over $$\mathbb{Q}$$ and $$Z(D)=\mathbb{Q}$$;

• $$K:=\mathbb{Q}(2^{1/3})$$ is a maximal sub-field of $$D$$?

My way towards solution: if $$D$$ is such algebra, then consider $$x\in D$$ outside $$K:=\mathbb{Q}(2^{1/3})$$. If conjugation by $$x$$ leaves $$K$$ invariant then it induces an automorphism of $$K$$ which fixed $$\mathbb{Q}$$; the only possibility of this is trivial automorphism, which means $$x$$ centralizes $$K$$, contradiction.

In fact, we can show that $$D=K\oplus xK \oplus x^2K$$ as a vector space.

Further $$x^3$$ centralize all generators of $$D$$, so $$x^3\in\mathbb{Q}\setminus \{0\}$$. Next, how should I proceed to determine structure of $$D$$?

I never studied division algebras other than quaternions and fields. I do not know how this question will be, but I was trying to see whether after $$2^2$$, can we get non-commutative division algebra whose dimension over its center is $$3^2$$? So a simple case I thought is through above questions, I was unable to complete the solution of existence.

## Can I use algebra boolean to reduce the number of lines in my code?

I am recently studding computer science and I was introduced in algebra boolean. It seems that algebra bool is used to simplify logic gates in hardware in order to make the circuit design minimal and thus cheaper. Is there any similar way that you can use it to reduce the number of codes in your software in higher level like C++, c# or any other higher languages?

## Proving that the exterior algebra is symmetric by looking at the polynomial ring

Recall that a finite dimensional algebra $$A$$ over a field $$K$$ is called Frobenius in case $$A \cong D(A)$$ as right modules, and it is called symmetric in case $$A \cong D(A)$$ as bimodules (where $$D=Hom_K(-,K)$$).

It is known that the Koszul dual algebra $$A^{!}$$ of an Artin-Schelter regular Koszulalgebra $$A$$ is a finite dimensional Frobenius algebra.

Question 1: In case $$A=K[x_1,…,x_n]$$ is the polnyomial ring in $$n$$ variables over a field $$K$$ of characteristic 0, $$A^{!}$$ is an exterior algebra and in case I made no mistake this should be symmetric if and only if $$n$$ is odd. My question is whether one can prove this just by looking at $$K[x_1,…,x_n]$$. Or asked differently: What property corresponds to being symmetric in $$K[x_1,…,x_n]$$, if there is any? (so $$K[x_1,…,x_n]$$ should have this property only for odd $$n$$)

Question 2 (more general): Is there a condition on $$A$$ that describes when $$A^{!}$$ is a symmetric algebra?

## Order in quaternion algebra for Fuchsian group

It may be that I am missing something very simple. In S. Katok’s book “Fuchsian Groups”, Lemma 5.3.3, we have the following.

Lemma: Let $$\Gamma$$ be a Fuchsian group of finite covolume, $$k_0=\mathbb{Q}(\mathrm{tr}(\Gamma))$$. Assume that $$[k_0:\mathbb{Q}]<\infty$$ and $$\mathrm{tr}(\Gamma)\subset \mathcal{O}_{k_0}$$ (the ring of integers of $$k_0$$). Then, $$\mathcal{O}_{k_0}[\Gamma]=\left\{\sum_i a_i \gamma_i\,:\, a_i\in\mathcal{O}_{k_0},\,\gamma_0\in\Gamma\right\}$$ is an order of the quaternion algebra $$k_0[\Gamma]=\left\{\sum_i a_i\gamma_i\,:\, a_i\in k_0,\,\gamma_0\in\Gamma\right\}.$$

In the proof, we have the following argument.

We may assume that $$\gamma_0=\begin{pmatrix}\lambda\&\lambda^{-1}\end{pmatrix}$$, $$\lambda\neq 1$$ and $$\Gamma\subseteq \mathrm{PSL}(2,K_0)$$ where $$K_0=k_0(\lambda)$$. If $$\gamma=\begin{pmatrix}a&b\c&d\end{pmatrix}\in\mathcal{O}_{k_0}[\Gamma]$$, then $$a+d$$ and $$\lambda a+\lambda^{-1}d$$ are in $$\mathcal{O}_{k_0}$$. Notice that $$\lambda$$ and $$\lambda^{-1}$$ are units in $$K_0$$ and $$\mathcal{O}_{k_0}$$ is a subring of the ring of integers of $$K_0$$, so $$a$$ and $$d$$ are in the fractional ideal $$\frac{1}{\lambda^2-1} \mathcal{O}_{k_0}$$ of $$K_0$$.

I don’t understand why $$a$$ is in that fractional ideal. It seemed to me that $$(\lambda^2-1)a=\lambda(\lambda a+\lambda^{-1}d)-(a+d)\in (\lambda-1)\mathcal{O}_{k_0}-\mathcal{O}_{k_0}.$$ But I don’t see why that should lie in $$\mathcal{O}_{k_0}$$. Also, why is that ideal fractional? Wouldn’t that imply that $$(\lambda^2-1)\in \mathcal{O}_{k_0}$$?

On the author’s website, there is a mention of this Lemma in the errata: http://www.personal.psu.edu/sxk37/errata.pdf but it just says: “The end of the proof of Lemma 5.3.3 was modified according to M.Katzโs suggestion.”

Thanks for any clarification!

## An example of a Banach algebra with a specified property

I asked this question (https://math.stackexchange.com/questions/3076735/an-example-of-a-banach-algebra-satisfying-given-conditions) but unfortunately no one answered it. Please help me to find an example of a Banach algebra ( if any) with the following property:

Non-commutative non-unital Banach algebra $$A$$ for which $$aa_0 -a_{0}a$$ lies in the annihilator of $$A$$ for any $$a\in A$$.

Here $$a_0$$ is an element of $$A$$ not belonging to its centre $$Z(A)$$.

Could you please suggest me a good reference (on Banach algebras) including examples like this?

Any help is appreciated.

## Conditional expectation with respect to a sigma algebra

As part of a more complex problem I have this passage: $$E(X_{n+1}-X_{n}|F_{n})=E(X_{n+1}-X_{n})$$ where $$X_{n}$$ is a random variable defined as $$X_n(w)= \begin{cases} 0 \quad \frac{1}{n} < w \leq 1 \ n-n^2w \quad 0 \leq w \leq \frac{1}{n} \end{cases}$$ and $$F_n=\sigma(X_{1},…,X_{n})$$ is the sigma algebra generated by $$X_{1},…,X_{n}$$

Now I understand intuitively why this is true but is it possible to prove it in a rigorous way? I know that I can use the law of iterated expectation to get away with it but I want to understand why $$X_{n+1}-X_{n}$$ is independent from $$F_{n}$$.

## Definition of Borel Field and Sigma Algebra

Are these terms equivalent?

• borel algebra
• borel sigma-algebra
• borel sigma-field
• a sigma-algebra generated by a topology
• borel field

(If a set is sigma-algebra, there exists a topology that generates the sigma-algebra, so any sigma-algebra is a borel sigma-algebra?)

By Chung’s A Course in Probability Theory, a borel field is defined as

• closed under complement
• closed under countable union
• closed under countable intersection

which looks like the definition of sigma-algebra. Are these also equivalent terms?

• borel field
• sigma-algebra
• sigma-field

According to http://mathworld.wolfram.com/BorelField.html, the definition of borel field does not require a set is closed under complement. Which definition is correct?

According to Is Borel-field different from $$\sigma$$-field?, are borel-* terms only used in terms of real space?