I have a doubt regarding the difference between "measures" and "pre-measures". On an exercise I am asked to show that a family of sets $\Sigma$ is a sigma-algebra. And then, the the author asks "Is it possible to extend a pre-measure $\mu$ on $\Sigma$ to a measure?" measure on
but I've shown that $\Sigma$ is a sigma-algebra
in this case, isn't $\mu$ a measure?
I am asking why isn't $\mu:\Sigma \rightarrow \text{whatever}$ called a measure rather than pre-measure?
Hmm, yes @ÍgjøgnumMeg when considering the Petersen theorem which says that "every bridgeless, cubic graph contains a perfect matching" I was thinking whether it can be applied to cubic multigraphs as well
And also the theorem "for every bridgeless cubic graph, there is a 1-factor containing any specific edge"?
Do you think we can apply it to any cubic graph, whether simple or not @ÍgjøgnumMeg?
Ok, your comment is very useful, Thank you very much @ÍgjøgnumMeg . Another small question, suppose I draw a Cayley graph of a group G with respect to a generating set S, by thinking that there exists an edge from $g$ to $gs$ for $g \in G, s \in S$, and another Cayley graph, by thinking that there is an edge from $g$ to $sg$ for $g \in G, s \in S$. i.e. I consider left action of G on S in one Cayley graph, and right action of G on S, in the other.
Then if i erase the labels of the vertices of both Cayley graphs and consider the underlying undirected Cayley graphs, they both are the same, right?
As can be seen in math.stackexchange.com/questions/3338047/… , no poster actually addressed my question, yet some posters had no problem to use their voting power in order to negatively score it.
yeah. Doesn't really help my approach but important to get it right
Let's try the matching pennies case. Then $a^\top P b = (a_1 -a_2)(b_1-b_2)$
If $a_2>a_1$, then to minimize this you'd pick $b_1=0,b_2=1$ and get $a_1-a_2<0$. If $a_1>a_2$, then to minimize this you'd pick $b_2=1,b_1=0$ to get $a_2-a_1<0$. If $a_1=a_2$, then $a^\top P b =0$ and it doesn't matter what $b_1,b_2$ are
The key bit of note here with respect to my strategy, though, is the sudden switch from (0,1) to (1,0) as you go from $>$ to $<$. so you're not looking at a critical point
you're minimizing a linear function, so you look at the boundary
I'm not sure how helpful that is is when $n>2$, though. For instance, if $n=3$ then the set of allowed $b$ is the triangle formed by (1,0,0), (0,1,0), (0,0,1)
okay, i think the idea is this. since you're minimizing a linear function, it must be the case that the minimum in $b$ is achieved at one of its vertices ($b=e_k$)
in which case the minimum $w$ is one of $a^\top P e_k$. Therefore one has $w\leq a^\top P e_k$ for $k=1$ to $n$
We don't know what $w$ is yet, but it can't violate any of those bounds
And then we try to make $w$ as big as possible, subject to those constraints.
In the case above, that gives us the following. We have $a^\top P e_1=a_1-a_2$ and $a^\top P e_2=a_2-a_1$. So whatever the minimum w/r/t $b$ is, it must satisfy $w\leq a_1-a_2$ and $w\leq a_2-a_1$. This is empty unless $a_2=a_1$
the hard part of the above sentences is probably the "therefore"
Here is something that has been bugging me for a few days now. Let $\mu$ be a Radon probability measure on a topological space $X$. Suppose that $\mathrm{supp}(\mu)$ is discrete, is $\mu$ a sum of weighted Dirac measures centered at points of the support?
Note that this is false in general, the Dieudonne measure on $[0,\omega_1]$ is a probability measure supported on a point which is not a weighted Dirac, however it isn't inner regular so it isn't Radon either
whats the generalisation to higher dimensions of the statement that if a continuously differentiable function $f:\Bbb R\to\Bbb R$ is zero at some point $a$ there exists a continuous function $g$ with $f(x)=(x-a) g(x)$ called?
Do you know how the reduced group $C^\ast$-algebra is defined? (I'm just dealing with discrete groups, no Haar measure on locally compact groups technicalities)
@user76284 Do you have $(x,x)\in\varnothing$ for all $x\in X$, where $X$ is the set the relation is defined on?
Speaking of empty domains, is axiomatizing free, inclusive logics more difficult than axiomatizing classical logic? Is that why they're less commonly used?
I want to see why $C^\ast_\lambda(G)$ is isomorphic to $C^\ast_\rho(G)$ (the former uses the left regular representation, the latter is the same construction with the right regular representation)
I think the map $e_{g}\mapsto e_{g^{-1}}$ should give a multiplicative morphism between these two algebras (and the other parts of $*$-morphism are no difficulty)
Maybe we are talking about different things,I was expecting:
$C_\lambda^*(G)$ to be the completion of $\Bbb C[G]$ where the multiplication is $(z_1g)\cdot (z_2h) = z_1 \lambda(g)(z_2h)$ and $C_\rho$ the same but with $\lambda$ replaced by $\rho$
$U:\ell^2(G)\to \ell^2(G)$, induced by $e_g\mapsto e_{g^{-1}}$ is an isomorphism that squares to identity. And $U\lambda(g)U(e_h) = e_{hg^{-1}}= \rho(g)e_h$
i guess you could define $R_g(e_h) = e_{hg}$ and then $\rho(g)=(R_g)^*$, but I think you might get $\lambda(g)$ like that
$\langle e_h , R_g e_f\rangle = \delta_{h,fg} = \delta_{hg^{-1},f} = \langle R_{g^{-1}} e_h, e_f\rangle$ so actually $(R_g)^* = R_{g^{-1}}$ (for groups) so maybe that is the construction
For semigroups $(R_g)^\ast$ is the map that on $gG$ looks like $gx\mapsto x$ and it is zero otherwise (writing $g$ for $e_g$), so it is kind of an inverse
I had used $R_g$ just as the "right multiplication map" on $\ell^2(G)$, i think this is a different map than you have written (coincidence of notation)
Hmmm I'm confused, this notes suggest that passing to the opposite semigroup turns the left regular representation into the right regular one, supporting the idea that $\rho(g)(e_h)=hg$
Well, using $\rho(g)(e_h)=hg$ or $\rho(g)=(R_g)^\ast$ makes no difference if we only care about the C*-subalgebra of $B(\ell^2(G))$ that they generate
This is silly, I found a ton of papers talking about the right regular representation of a semigroup and not a single source defining it
