Hello, does anybody know where I can find a derivation of the following formula (or its name so I can look it up myself): $$\exp(A) H \exp(-A) = \sum_{n=0}^\infty \frac{1}{n!}\mathrm{ad}^n_{A}(H)$$ or written otherwise $$\exp(A)H\exp(-A) = \exp(\mathrm{ad}_A)(H)$$ here $A, H$ are operators in some $C^*$ algebra (probably Banach also works), although I'm sure there is a $H$ in Lie group $A$ in Lie algebra version of the statement
@KaumudiHarikumar you can just ask your question, don't "ask to ask" as it says in the side bar^
@Maks I can't stick around to answer this, but I'm also not sure I understand what's being asked. Are you saying, what is the mirror image of (1,2,3) with respect to the plane created by (0,1,1) and (1,0,1)? If so, that's probably a clearer statement than the above.
What can you do the composition (I'll use this circle charecter for now " 。") operation between? A function and a value seems reasonable, for example: Increment 。4 = 5
I'm looking at stuff to do with groups and it seems that it can also be done between two states, the values themselves that go into sets
This is a lot like applying matrices to each other, which seems to mean "dot product", like you just multiply them together. I'm increasingly thinking of applying values to other values as being multiplication, is this reasonable?
let's suppose x^2 - 5x +6=0, if u put the value of 3 or 2 it will proof the equation. but that doesn't mean that only 3 or only 2 could be the value of x. cause both 3 and 2 are the solutions @Obliv
that's what i want to say @Obliv , it is very possilble that those |AB| / |CD| = m , coincedently.... i know this given problem is true... or else no one would use this for so many years.... but i just wanted a solution if u want to give.
@ffahim It's certain that if you define m as the ratio of the two vectors, then multiplying one vector by the other (using the correct ratio that is) will leave you with the same vector. This is logically sound. There is nothing to prove.
If these vectors are parallel, then their directions are the same (or one is pointing in the opposite direction. In this case m must be negative). The way we define vectors is by both magnitude and direction. If they are both equivalent, then there is no way to tell AB from CD that is why we call them equal.
A conformal map is holomorphic. But, (at least intuitively) if a map is conformal (preserves angles), it has inverse, which preserves angles too... But a conformal map is (nonconstant) holomorphic!
Is there a name for a matrix that is completely filled with nonzero values? One has names for things such as diagonal or tridiagonal matrices, but is there something for the filled one?
I think of "generic" as "element picked randomly from a (Zariski) open set". Matrices with at least one zero entry indeed form a closed set in the space of all matrices; hence matrices with nonzero entries form a Zariski open set. So that's a fine choice of word depending on how one interprets it.
There are certainly not much matrices with a single zero entry out there so I don't think there's any way to misinterpret it.
@AndersonFelipeViveiros What's your definition of conformal? Holomorphic maps which have nonzero derivative at a point preserve angles, but clearly such maps need not have a global holomorphic inverse.
@TedShifrin So I learned about that holomorphic sections of the line bundle $\mathcal O(-D)$ associated to the effective divisors should be seen as holomorphic functions whose divisors multiply $D$, i.e. $(f)\geq D$.
This fits nicely into the short exact sequence of sheaves (note: I use the same symbol for a vector bundle and its sheaf of sections) $$ 0\to \mathcal O(-D) \to \mathcal O_X \to \mathcal O_D \to 0$$ which *defines* $\mathcal O_D$ in Huybrechts
Now, I'm trying to think what happens for non-effective (arbitrary) divisors
Is it as simple as
$$ 0 \to \mathcal O(-D) \to \mathcal K_X \to \mathcal O_D \to 0$$ where $\mathcal K_X$ is the sheaf of meromorphic functions?
@AndersonFelipeViveiros Conformal maps are locally injective, but not necessarily globally injective. You could have a map from the disc to an annulus that "loops around" the annulus, and isn't injective.
@Danu: BTW, $\mathscr O(-D)$ makes sense even when $D$ isn't effective, as you know. It just is the sheaf of germs of meromorphic functions with $(f)+D\ge 0$. So you're worrying that this is not a subsheaf of $\mathscr O$ when $D$ fails to be effective.
OK, so you need the quotient to have principal parts at each of the poles (up to the prescribed order), @Danu. For example if $D=nP$, we should have $$0\to \mathscr O\to \mathscr O(nP) \to \Bbb C_P^n\to 0$$
Maybe you'd prefer I say "Use the sheaf exact sequence Z -> R -> S^1". The latter is H^0(X,S^1). The sheaf of continuous R-valued functions is acyclic.
@Danu: So, in general, I think you'd need to split the divisor up as $D-E$, where $D$ and $E$ are each effective. I was just writing $\Bbb C^n$ in a way as to emphasize you think of one copy of $\Bbb C$ for each Laurent coefficient.
@MikeMiller an analogous version of the geometric interpretation of $H^1(X; \Bbb Z/2) \cong [X, \Bbb{RP}^\infty]$, I mean to say. Can that be done (something something principal bundle)? The interpretation of the $\Bbb Z$ coeff version I know is by making a map $X \to S^1$ transverse to a pt and taking preimage.
@TedShifrin I don't believe that. How else would he understand a "bundle-theoretic proof" that RP^infty = H^1(X;Z/2)? To show that the latter corresponds also to line bundles, you want the sheaf exact sequence.
The "bundle-theoretic proof" you're providing is that RP^infty = K(Z/2,1), which classifies both line bundles and elements of H^1(X;Z/2). But that doesn't pass through line bundles to prove the isomorphism between [X,RP^infty] and H^1.
A proof that doesn't start with that isomorphism via homotopy theory uses sheaves.
Not on a compact RS, @Danu. The only global holomorphic functions are constants, and if you require that the functions vanish somewhere, you're dead. You need to allow poles to get global meromorphic functions.
No, that's not right! I have a proposition that says that for any effective divisor, there exists a nonzero global section of $\mathcal O(D)$ such that $Z(s)=D$.
Fine. [X,S^1] = [X,BZ] classifies principal Z-bundles. Becauze Z is discrete, these are classifies by homomorphisms pi_1 -> Z.
Not very exciting.
If you wanted to classify what subgroups of a given group arose as centralizers, how would you approach that?
The obvious person to ask is @arctictern. I'd like to do this for SO(3) and SU(2) (closed subgroups, of course). I know the answer and the geometric meaning, but not the proof.
Too early for martinis, and I'm playing bridge tonight with a good player, so I need to be awake. Plus, my stomach is totally messed up from this antibiotic, so no booze.
We'll see, @Balarka: This is the type that's normally prescribed for these sorts of issues, so I dunno. I had it years ago without the side-effects ... I'm just tooooo old.
There's so many successful and reasonably honest capitalists that would probably make decent presidents. The one running now is neither of those things.
how this could be proofed that if AB and CD are two non zero vectors then AB =mCD IF AND ONLY IF AB II CD.? how this could be proofed ?
