TITLE: Adjoint functors preserve (co)products
QUESTION [5 upvotes]: Let $F:\mathscr{A} \to \mathscr{C}$ and $G:\mathscr{C} \to \mathscr{A}$ be an adjoint pair of functors.
I am trying to show $G$ preserves products  and $F$ preserves coproducts.
So to start we have a bijection
$\tau_{AC}:\operatorname{Hom}(FA,C) \to \operatorname{Hom}(A,GC)$
as well as the normal naturality diagram for both variables.
It suffices to prove only one pair of the question - the other will follow form dualising the proof (or I am guessing working in the opposite categories or some other nice trick).
I decided to try show that $G$ preserves products. So we have the usual diagram for the product: i.e. given $C_1,C_2 \in \mathscr{C}$ there exists morphisms $g_1:C_1 \times C_2 \to C_1$ and $g_2:C_1 \times C_2 \to C_2$ such that for any other map $X \to C_i,i=1,2$ there is a unique map $\theta:X \to C_1 \times C_2$.
Now I can immediately apply the functor $G$ to this. This doesn't quite give what I want. In particular it gives $G(C_1 \times C_2)$ in the usual universal diagram. To solve the problem we need to replace that with $G(C_1) \times G(C_2)$.
I've played around for quite awhile to no real avail. Obviously I need to use naturality. So given $g:C \to C'$ in $\mathscr{C}$ we have that
$(Gg)_* \tau_{AC} = \tau_{AC'} g_*$.
where $$g_*:\operatorname{Hom}(FA,C) \to \operatorname{Hom}(FA,C')$$ and $$(Gg)_*: \operatorname{Hom}(A,GC) \to \operatorname{Hom}(A,GC').$$
I was thinking the way to go would be to use the maps $g_i: C_1 \times C_2 \to C_i$ and play around with the natural transformation diagram to get something nice, but I keep getting nowhere.
Any hints?
(I would add that I am aware there is a more general result that $F$ preserves colimits and $G$ preserves limits, but I am not familiar enough with them to work with them just yet.)

REPLY [1 votes]: Suppose we have an adjunction as in your question $F \dashv G $, and a product $A\times B$ with projections $\pi_1 : A\times B\rightarrow A$ and $\pi_1 : A\times B\rightarrow B$. If we want to show $G$ preserves the product $A\times B$ we need only show that $G(A\times B)$ with the projections maps $G(\pi_1) : G(A\times B)\rightarrow G(A)$ and $G(\pi_1) : G(A\times B)\rightarrow G(B)$ has the universal property that the product $G(A) \times G(B)$ must have (if it exists) and by uniqueness they are the same.
Given maps $c_1 : C\rightarrow G(A)$ and $ c_2 : C\rightarrow G(B)$ via the adjunction we obtain maps $\bar c_1 :F(C)\rightarrow A$ and $\bar c_2 :F(C)\rightarrow B$ which, by the universal property of $A\times B$, induce a unique map $\bar u:F(C)\to A\times B$ such that $\bar c_i = \pi_i \bar u$ (i = 1,2). Going back through the adjunction we get a map $u: C \to G(A \times B)$ such that $c_i = G(\pi_i) u$ (i = 1,2) by naturality of the adjunction, and it is unique among such maps as the adjunction is a bijection of hom-sets so $G(A\times B)$ has the universal property that the product $G(A)\times G(B)$ must have so $G(A\times B) = G(A)\times G(B)$ hence $G$ preserves any products which exists in $\mathcal C$. Note that this argument easily generalizes to show that $G$ preserves all limits which exist in $\mathcal C$


TITLE: Floor and Ceiling function
QUESTION [0 upvotes]: prove that x is a real number then $\lfloor −x \rfloor = −\lceil x\rceil$ and $\lceil −x \rceil = −\lfloor x \rfloor$. If I were to put a real number say like $1$, wouldn't it be right? But if I put a value like $1.5$ it would be incorrect? I don't think this is the way I should be proving it though so could anyone help me out.

REPLY [1 votes]: HINT: As you noted, it’s easy to see that the identities hold when $x$ is an integer. Suppose that $x$ is not an integer; then there is an integer $n$ such that $n<x<n+1$. Multiplying this by $-1$, we see that $-(n+1)<-x<-n$. What are $\lfloor x\rfloor$, $\lceil x\rceil$, $\lfloor-x\rfloor$, and $\lceil-x\rceil$ in terms of $n$? Do the identities hold?
If you think in visual or physical terms, it might be helpful to visualize multiplying the number line by $-1$ as rotating it $180$° about the point $0$, so that the numbers $x$ and $-x$ change places. If $x$ is trapped between two integers, it will still be trapped between their negatives after the rotation, but the order of the three numbers will have been reversed.


TITLE: "Fictive" irreps of the enveloping general Lie algebra
QUESTION [0 upvotes]: Notation abuse warning: I will use the E7 series irrep names. You'll soon see why.
In the general Lie algebra, the irrep you "start" with is $J$, the
adjoint. From the series for $J\bigotimes{J}=1+J+A...$ ($1$ is the 1-dimensional irrep) you get $A$, the antisymmetric one. (The quantum dimensions up to here are all given e.g. in Westbury's "R-matrices and the magic square" paper.) The next step would be $J\bigotimes{A}=J+A+S...$ where you can "fish out" the symmetric irrep $S$ from the Clebsch Gordan series. (Does anybody have the quantum dimension $Dim(S)=S(\alpha,\beta,\gamma)$ handy, just as for $J$ and $A$? In principle I could compute it myself, but this would speed up my work considerably.)
Now comes the silly part :-) In the E7 series, $V\bigotimes{V}=1+J+S+A$ and I can now backwards compute the dimension of the defining irrep $V$ and see what comes out for, say, E8. Well, nonsense is coming out, which is hardly surprising. Still, I would have expected
integer nonsense or even better a zero ($324-273-52+1=0$ for F4 -the minus signs are only very mild cheating. :-)! Irrational nonsense is baaaad.
So: Is it possible to bring all irreps, even from different rows of the magic square, in an 1:1 correspondence (if neccessary by inventing fictive irreps which still have integer dimension) or am I completely barking up the wrong Lie?

REPLY [1 votes]: The (quantum) dimensions are all given in Vogel's papers (all but one unpublished). There may be some errors in my paper you refer to as I confused Vogel's parameters with the values of the quadratic Casimir.
I don't follow your "silly part". The representation $V$ is not defined for $E_8$ and I don't see how you get a proposed quantum dimension.


TITLE: Find the argument of $ \frac{-1 + \sqrt3 i}{2+2i} $
QUESTION [2 upvotes]: I rewrite equation $ \frac{-1 + \sqrt3 i}{2+2i} $ as $$ \frac{ \sqrt3 - 1}{4} + \frac{ \sqrt3 + 1}{4} i  $$ using the conjugacy technique.
And set forward to find the argument of this complex function.
I'm assuming the argument is the angle between the positive real-axis and the line?
I do this by trying to calculate:
$$ \mathbf{cos}^{-1}  \frac{ \sqrt3 + 1}{\sqrt3 - 1} $$
But my calculator doesn't seem to understand what I'm saying. Is this ratio just totally wrong?
Help please

REPLY [1 votes]: using eulers equation$$e^{i\theta} =\cos \theta+i\sin \theta$$the given expression can be written as $$\frac{\frac{-1}{2}+\frac{\sqrt{3}}{2}i}{1+i}$$which is$$\frac{e^{i\frac{2\pi}{3}}}{ e^{i\frac{\pi}{4}}}$$you get the resulting arguement$\frac{5\pi}{12}$


TITLE: Maximum-Value Secretary Problem
QUESTION [4 upvotes]: Background:
The classic secretary problem has the simple solution of rejecting the first 1/e applicants and then selecting anyone who was better than the best in the rejected set.  However, in the real world secretaries are not of value 0 if they are not the best, and so choosing the 2nd best is normally far better than choosing the worst.  Also, if you stick to the classical approach, there's a 1/e chance that you will receive a random secretary from the set that excludes the best secretary.
Real-World Example:
Say that you've interviewed 98 out of 100 secretaries and have not found any better than secretary #32.  You then interview the 99th secretary and find that she is second only to #32.
Analysis:
Under the classical problem, you'd reject #99 as they have a value of 0 (not the best).  Thus, you'd take a random secretary in favor of the 98th or 99th best, depending on whether the last secretary is the best or not.
However, the classical solution would not be the best decision in nearly all real-world cases, as you are taking a random applicant instead of one better than at least 97/99 others.
Question:
What is the proper stopping strategy to maximize the expected-value of the secretary you hire?

You have n secretaries.
Each secretary has a linear value assigned after each interview (a secretary ranked 4 is assumed to be twice as valuable as one ranked 2).
The value-distribution of secretaries is unknown.
You must accept or reject each secretary immediately following their interview with no recalls.
You must maximize the expected value of the secretary you hire.

REPLY [0 votes]: Let us first analyze the expected value of the candidate chosen by the same stratey, which maximizes the probability of hiring the best candidate. As you mentioned, the strategy is the following: Rejecting the first $n/e$ candidates and memorizing the best candidate $c^{relativeMax}$ amonst them. We then select the first candidate who is better than $c^{relativeMax}$.
Let's call the strategy we just described $S$, the sequence of candidates it receives $\sigma$ and $w_S(\sigma)$ defined as the weight or value of the candidate which strategy $S$ chooses when given $\sigma$. And let $c^*$ represent the best candidate.
Then, we can conclude
$$E[w_S(\sigma)] = \sum_{c\in\sigma} P[\text{candidate c is selected}]\cdot w(c)
 \ge P[c^* \text{is selected}] \cdot w(c^*) \ge \frac{1}{e}\cdot w(c^*).$$
[On page 15, Corollary 2.2 of this thesis, you can see the proof:
http://darwin.bth.rwth-aachen.de/opus3/volltexte/2014/5270/pdf/5270.pdf]
So, our the expected value of the chosen candidate of our strategy is at least $1/e$ of the best candidate.
The only point left to prove is, that there is no strategy, which can obtain a candidate with an expected value better than $1/e \cdot w(c^*)$. More formally, we have to show that:
$$\not\exists  \text{ Strategy } S: \exists r < e:  \forall \sigma: E[w_S(\sigma)]\geq \frac{1}{r}.$$
Once we have shown this, we know we cannot do better than $1/e \cdot w(c^*)$ for the expected payoff and our intial strategy also is not only a strategy for maximizing the probility of the best candiadate, but also a strategy for maximizing the expected value of the selected candidate.
This has also been proven in the thesis I mentioned above (Corollarly 2.4), but it contains a logical mistake assuming, that one instance for $\sigma$ is enough to proove this. For my bachelor thesis, I am working on this proof and once I have completed it, I will attach is to this post.