Grothendieck group of the category of boundary conditions of topological field theory

In this paper also the journal front page, eq. 2.14, it introduces the
the Grothendieck group $K^0$ of the category of boundary conditions of topological field theory.

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My question is that

  • what exactly is the Grothendieck group $K^0$ of the category of boundary conditions of topological field theory really mean?
  • Does this say that the boundary conditions of topological field theory can be related or classified by a Grothendieck group? Is this an abelian group or nonabelian group? ($K^0(BCondμ(M^{d−1}) )=?$)
  • Should the boundary conditions of topological field theory classified by some bimodule of certain modular tensor category? how is this related to a Grothendieck group?

complexity theory – Functions with small support have small circuits

I have been trying to understand the use of circuit models for boolean functions, and came across this question, that I am trying to struggle to understand:

Show that if a function $fcolon {0,1}^n→{0,1}$ has a support of size $k$, then it can be decided by a circuit of size $O(nk)$.

I understand that for any binary functions of this sort, the maximal size of the circuit is $O(n2^n)$, and I have a feeling that this could be useful for proving the statement, but I am not exactly sure where to proceed from there. Furthermore, I also understand that the “support” referred to in this question is actually the language defined by the function, but again, this leads me nowhere.

Any help on this would be much appreciated.

linear algebra – Are there any results in generalizing matrix theory to multidimensional arrays?

In matrix theory(2-dimensional arrays), we can define addition, multiplication, rank and determination etc. I’m working on generalizing these properties to multidimensional arrays as many as possible. Are there any results in this way? I’d really appreciate it if you could provide some references.

group theory – Endomorphism That Squares an Element

My professor challenged me to solve the following problem:

Come up with a finite group, $G$ and endomorphism $f$ such that the fraction of elements, $x in G$, satisfying $f(x) = x^2$ is maximized (and not equal to $1$).

So far, the biggest I’ve seen is the dihedral group, $D_4$, with endomorphism $f(x) = e$, where $e$ is the identity element. In that case, $frac{3}{4}$ of the elements work (namely $e, r_2, s_1, s_2, s_3, s_4$).

However, I am confident that larger fractions exist.

So far, the main idea that I have come up with is that such a group CANNOT be a direct product. That is because if $a$% of the elements in Group $G$ with endomorphism $f$ satisfy $f(x) = x^2$ and $b$% of the elements of the elements in Group $H$ with endomorphism $h$ satisfy $f(y) = y^2$, then then $x$% $cdot y$% of the elements in $G times H$ with endomorphim $f times g$ will work.

Unless $x$ or $y$ is $100$%, this fraction will only decrease after a direct product. Even then, taking the direct product will not increase the fraction of elements that work – it will only keep it the same.

Some other points that I have thought of is that for all of the dihedral groups, $D_4$ gives the optimal fraction. Additionally, none of the $S_n$ (symmetric) groups can give this high of a fraction (I don’t think at least).

Does anyone know a way to find the largest possible fraction that works (and hopefully, know a way to prove that is actually the maximal value)?

probability theory – Deterministic algorithms for computational distance between distributions

Computational distance between sequences of distributions ${X_i}_{i in mathbb{N}}$ and ${Y_i}_{i in mathbb{N}}$ can be defined as the maximum, over all probabilistic polynomial time algorithms $A$, of
$$ left|underset{x sim X_{n}}{mathsf{Pr}}(A(x) = 1) – underset{x sim Y_{n}}{mathsf{Pr}}(A(x) = 1)right|, $$

for each $n in mathbb{N}$. We can, very similarly, define computational indistinguishability to be when the computational distance is a negligible function of $n$. Here is my question: can we use a deterministic polynomial-time algorithm in place of a probabilistic one without loss of generality? Can we say that the maximum computational distance is achieved by a deterministic algorithm? Will this algorithm be non-uniform?

nt.number theory – Explicit bound for sum of Kloosterman sums

What are the best fully explicit upper bounds one can give for the sum
$$leftlvert sum_{n=N}^{infty} frac{S(a,b;n)}{n} ,I_1!left(frac{4 pi sqrt{|ab|}}{n}right) rightlvert$$
where $S(a,b;n) = sum_{gcd(k,n)=1} e^{2 pi i (ak+bk’)/n}$ is the ordinary Kloosterman sum?

As a generalization one may consider any sufficiently regular function $f(C/n)$ instead of the Bessel function.

A trivial bound follows from using Weil’s bound $|S(a,b;n)| le tau(n) sqrt{gcd(a,b,n)} sqrt{n}$, plugging in an explicit bound for $tau(n)$, and summing the series (see for example Brisebarre and Philibert), but this is quite pessimistic.

The problem of bounding such sums is discussed in Sarnak and Tsimerman, On Linnik and Selberg’s conjecture about sums of Kloosterman sums. As far as I can tell, there are no published bounds (other than the trivial one) with fully explicit constants.

operator theory – I don’t understand this. What is it doing?

Let $A$ be a separable C*-algebra and ${pi_n}_{ninmathbb N}$ be a sequence of finite dimensional representations of $A$ such that for every $aneq 0$ there is $n$ such that $pi_n(a)neq 0$. In other words, $bigoplus_n pi_n $ is faithful. (such $A$ is said to be residually finite dimensional.)

Let $B$ be unital separable and ${pi_n}_{ninmathbb N}$ be such sequence. Assume every $pi_n$ is irreducible and has dimension $k(n)$. Also assume each representation in ${pi_n}_{ninmathbb N}$ repeats infinitely many times.

Suppose $l:mathbb Nto mathbb N$ is an increasing sequence.

For each $n$, define $phi_n:M_{k(n)}to M_{k(n)}(B)$ by viewing $M_{k(n)}$ as a subalgebra of $M_{k(n)}(B)$, since $B$ is unital. Then define $psi_n^1=phi_ncirc pi_n:Bto M_{k(n)}(B)$.

By defining $h_1:bmapsto text{diag}(b,psi_1^1(b),…,psi_{l(1)}^1(b))$
, $h_1$ is a homomorphism from $B=M_1(B)=M_{I(1)}(B)$ to $M_{I(2)}(B)$, where $I(1)=1$ and $I(2)=1+k(1)+…+k(l(1))$.

Suppose $h_m:M_{I(1)}(B)to M_{I(2)}(B)$ and $psi_n^m:M_{I(m)}(B)to M_{I(m)k(n)}(B)$ are defined. Define $psi_n^{m+1}=psi_n^1otimes 1_{I(m+1)}:M_{I(m+1)}(B)to M_{k(n)I(m+1)}(B)$ and define $h_{m+1}:bmapsto text{diag}(b,psi_1^{m+1}(b),…,psi_{l(m+1)}^{m+1}(b))$ as homomorphism from $M_{I(m+1)}(B)$ to $M_{I(m+2)}(B)$, where $I(m+2)=I(m+1)(1+k(1)+…+k(l(m+1)))$.

Then we have unital homomorphisms $h_m:M_{I(m)}(B)to M_{I(m+1)}(B)$ and therefore we can define the inductive limit C*-algebra $A=lim_{ntoinfty}M_{I(m)}(B)$.


Proposition. $A$ is simple.


This is from An introduction to the classification of amenable C*-algebras, p158. This definition is too complicated for me to understand.

Since $h_m$ maps $b$ to the matrix with $b$ on its diagonal, $h_m$ is a norm preserving injection. Therefore there are $M_{k(n)}(B)simeq B_nleq A$ with $mathbf{closure}(bigcup B_n)=A$. It only needs to show there is simple $M_{I(n)}(B)$ for arbitrarily large $n$. However, the definition is so complicated that I don’t even know where to start.

As in the proof the book gives, it says,

For $0neq yin M_{l(n)}$, there is $pi_m$ such that $pi_motimes text{id}_{M_{I(n)}}(y)neq 0$. Let $n+lgeq m$, by considering $h_{n+l}circ h_{n+l-1}circ…circ h_{n+1}$ we may write $h_{n,n+l}(y)=H(y)oplus (pi_motimes text{id}_{M_{I(n)}}otimes text{id}_{M_J})$ for some positive integer $J$ and some homomorphism $H$. Since $psi_m^1otimes text{id}_{M_{I(n)}}otimes text{id}_{M_J}(y)$ is a nonzero element in $M_{I(n+l)}$, the ideal generated by $psi_m^1otimes text{id}_{M_{I(n)}}otimes text{id}_{M_J}(y)$ contains $M_{I(n+l)}$.

I do not understand it at all.

complexity theory – Configuration of a space bounded turing machine

A configuration of a turing machine is defined as the following:

an ordered triple (x, q, k) ∈ Σ∗ × K × N, where x denotes the string
on the tape, q denotes the machine’s current state, and k denotes the
position of the machine on the tape

I have read in a paper that a space bounded non-deterministic turing machine (NSPACE), has at most 2^(d*n) configurations on an input of length n, where d is a constant, how do we know that this is true? what is d? and how can we prove it?

rt.representation theory – Are there analogues of Young tableaux for more general group actions of the symmetric group?

Let $G = Sigma_n$ denote the symmetric group on $n$ elements. Let $underline{n} = {1,ldots,n}$. It is clear that $G$ acts transitively on $underline{n}$. Moreover, if we have a copy of $V$ ($V$ a finite-dimensional complex vector space) for every element in $underline{n}$, then we can form their tensor product

$$ bigotimes_{i in underline{n}} V $$

and it is clear that $G$ acts on it. This is related to the theory of Young tableaux and Young symmetrizers.

Now let $C_2(underline{n}) = { (i,j) in underline{n} times underline{n}; i neq j }$. $G$ also acts transitively on $C_2(underline{n})$. Now suppose also you have a copy of a finite-dimensional complex vector space $V$, one for each $(i,j) in C_2(underline{n})$, and you form their tensor product

$$ bigotimes_{(i,j) in C_2(underline{n})} V. $$

Then $G$ acts on that tensor product. Can one define analogues of Young tableaux in this case, which would give irreducible representations of $G$, which are subrepresentations of the “big” tensor product representation that I just wrote down?

I am mostly interested in the case where $V = mathbb{C}^2$ (but my questions make sense in a more general setting as well).

number theory – Can we determine which integers can be written as the difference of two cubes?

I would like to know whether we can determine which integers can be written as the difference of two cubes. Algebraically, this can be stated as finding the image of $f:mathbb{Z^2}to mathbb{Z}, f(x,y)=x^3-y^3$. I came up with this idea after I saw that any integer that is not congruent to $2$ modulo $4$ can be written as the difference of two squares. I tried to apply the same reasoning here, but to no result. I thought about looking at $x^3$ modulo $6$ and modulo $8$, but this doesn’t produce anything, because there are far too many combinations. So I wonder if this is doable or not.