if you have an ad-invariant inner product then $\mathfrak{g} = \mathfrak{h} \oplus Z(\mathfrak{g})$ where $\mathfrak{h}$ is centerless, so you can embed it in a matrix algebra by the little ad map, and the other half, the center, is abelian, so it independently embeds in a matrix algebra
then i just $\oplus$ them
This is already a large class, Lie algebras of all Lie groups isomorphic to $G \times \Bbb R^n$ where $G$ is compact
I was hoping semisimple is only a bit harder
in particular i was hoping if you add o to ad you get ado
Define the kernel of the evaluation-at-0 $\text{ev}_0 : C[0, 1] \to \Bbb R$ to be $C_0[0, 1]$, and equip it with uniform topology and the Borel $\sigma$-algebra thereof.
I can define a measure on the sets of the form $\{\gamma \in C_0[0, 1] : \gamma(t_1) \in U_1, \cdots, \gamma(t_k) \in U_k\}$ where $0 < t_0 < \cdots < t_k$ and $U_1, \cdots, U_k \subset \Bbb R$ open.
so you're trying to define a measure that controls paths on the line by how/when they travel between open sets? and that's somehow governed by Gaussians?
yeah you basically want to think of a BM as a random element on the space of paths
ah so the intuition is
you want a brownian motion to have iid normal increments, with variance equal to time-interval in the increment
so the probability that a bm should start at 0, halt at time t1 in U1, then halt at time t2 in U2, etc etc is (by markovian philosophy) probability that a bm should start at 0 then halt in U1 (on x1 say) at time t1, another totally independent bm should start at wherever the first guy halted in U1 (ie x1) then halt at time t2 - t1 somewhere in U2 etc
$X : [0, \infty) \times \Omega \to \Bbb R$ be a stochastic process such that for some $\alpha, \beta, C > 0$, $\Bbb E|X_s - X_t|^\beta \leq C |t - s|^{1+\alpha}$. Let $\gamma < \alpha/\beta$.
There is a constant $K(\omega)$ such that $X_{-}(\omega)$ is locally $\gamma$-Holder continuous on the diadic rationals on $[0, 1]$
Let's see this. Define the event $G_n = \{|X_{i/2^n} - X_{(i-1)/2^n}| \leq 1/2^{\gamma n}\; \forall 0 < i \leq 2^n\}$.
which is in turn less than $C 2^{-n(1+\alpha)}$ by hypothesis. Combining, we get $\Bbb P(G_n^c) \leq C 2^{n(\beta \gamma - \alpha)} = C 2^{-n \lambda}$ where $\lambda = \alpha - \beta \gamma > 0$.
Clearly on $\bigcap_{n = N}^\infty G_n$, for all $p, q \in [0, 1]$ diadic rational we have $|X_p - X_q| \leq A(\gamma) |p - q|^\gamma$ whenever $|p - q| < 2^{-N}$, where $A(\gamma)$ is some constant depending entirely on $\gamma$
Now it's a Borel-Cantelli trick
$\sum_{n = 1}^\infty \Bbb P(G_n^c) = \sum C 2^{-n\lambda} < \infty$ so $\Bbb P(\limsup G_n^c) = 0$
so everything is eventually in $\bigcap_{n = N}^\infty G_n$ almost surely; which is what we want
This is apparently called the Kolmogorov continuity theorem. If there's some funky Holder assumption on the moments of the increments of the stochastic process then it's sample Holder-continuous on dyadic rationals (and thus the whole stochastic process can be modified to a sample-continuous process leaving distribution unchanged)
@Knight Here's a concrete exercise for you. Give me a differentiable function $f$ for which $\lim_{x\to\infty} f(x) = 0$ but for which $\lim_{x\to\infty} f'(x)$ does not exist.
@TedShifrin I’m sorry Ted, I tried hard but couldn’t find any of them. My train of thought was like this: a function having a limit zero as $x$ goes to infinity must be 1 over “some increasing function”
So, I tried $1/x$ , $\frac{1}{e^x}$ and $\frac{1}{ln x}$ and even their combinations but couldn’t get anything good
You see examples where some bounded, oscillating function, e.g. $\sin(x)$, is applied to some quickly increasing function, e.g. $x^2$, gives a function whose derivative goes wild.
@robjohn Sir you know, these days weather is so nice here. I really admire the cool weather caused by rain, dark clouds and green green floras
And I don’t know how but it reminds me of 9th Grade when our History teacher Mr.Chowfin was teaching French Revolution (it was the first Chapter) and whole class was quite and listening to his lecture, it was just very nice. I remember few names from French Revolution: Louis 16, Antoinette, Jacobian, Roberspiere, 3rd Estate and few more
@robjohn Are you working on preparing lectures for UCLA, In this nice weather where birds too are flying with freedom without any prohibitions by despotic sun ?
its funny i have not yet to see how all these change of basis tricks and calculations applied somewhere. I know physicists must use em alot.
having a basis for the space the eigenvectors ensures that the action of matrix does not rotate or do weird stuff with your axes which is good for visualization but it can still mess up the picture depending on what the matrix acts on.
Hey everyone, a quick question (sorry for the bad terminology...) - for abelian groups, when I see $\bigoplus_{i \in I} A_i$, should this mean the colimit of the discrete diagram ( discrete = no non-identity morphisms) (where this discrete diagram has $I$ many objects, even if $A_i = A_j$ etc.)
@robjohn Sir advice me on the issue of elementary integrals. I learned integration (substitution, by parts, partial fractions) two years ago. I want to master the elementary integrals, but whenever I pick any book they teach it from scratch and I cannot help leaving it.
My problem is that books like : Stewart Calculus, Thomas Calculus and Edward’s Calculus, the topics they teach I know them, all I want is to master those topics and I don’t know what to do. Suggest me something
Let $I$ be a discrete small category and $A\colon I\rightarrow C$ a diagram in $C$. Denote $A(i)=A_i$ and consider the coproduct $\coprod_{i\in I}A_i$ with the inclusion morphisms $\iota_i\colon A_i\rightarrow\coprod_{i\in I}A_i$ for all $i\in I$. This is, by definition, a co-cone over $A$. Another co-cone over $A$ consists of an object $N$ and morphisms $\varphi_i\colon A_i\rightarrow N$ for each $i\in I$. The universal property of the coproduct asserts there is a unique morphism $\psi\colon\coprod_{i\in I}A_i\rightarrow N$ such that $\psi\circ\iota_i=\varphi_i$ for all $i\in I$. This is p…
whether some $A_i$ are equal to one another or not isn't a concern, because everything is indexed over $I$ to begin with
To show how the convolution applied to an image can be viewed as matrix-vector multiplication, let's suppose that we want to apply a $3 \times 3$ kernel to a $4 \times 4$ input, with no padding and with unit stride.
Here's an illustration of this convolutional layer.
Now, let the kernel be de...
Jardine Simplicial Homotopy Theory page 163, first part of the first sentence: "Observe that $N_{0}A = A$, that ..." Am I missing something or is this wrong... It looks like $N_{0}A = ker{d_0}$
Very small for a question, but still couldn't figure it out since more than an hour: Jardine Simplicial Homotopy Theory page 163, first part of the first sentence: "Observe that $N_{0}A = A$, that ..." Am I missing something or is this wrong... It looks like $N_{0}A = ker{d_0}$
It's the standard tactic for deriving the value of this integral when we have infinite bounds on both ends, and my logic seems to check out, but with this derivation, I'm very leery of why it can give a correct answer (I've checked it several times for lots of conditions) and yet clearly cannot be used to find an antiderivative
Covering maps $\Sigma_g \to \Sigma_h$ are pretty restrictive; they exist iff $2 -2h$ divides $2 - 2g$, but in fact when this happens these coverings are realizable by rotation symmetries
$M_g/\Bbb Z_g$ is an orbifold, because $\Bbb Z_g$ action on $M_g$ is properly discontinuous (properly discontinuous = $U \cap gU \ne \varnothing$ for only finitely many $g$).
Non-oriented orbifolds just look like surfaces with boundary, where I think of the boundary as being a fold (since it's an orbifold point, not really a boundary point)
@BalarkaSen There's no classification of (g,n) and (g',n') for which this is allowed? I think there must be with a lot of work
There is no degree 4 branched cover of S^2 with 3 branch points of branching data (3, 1), (2, 2), (2, 2)
Even though that's an algebraically valid branching data
@feynhat Exercise: Give a novel definition of Euler characteristic of an oriented 2-orbifold and explain why it's interesting. Afterwards read the general definition from somewhere and then explain it to me.
I've noticed that there are a couple of things in mathematics that obey a type of "cut rule", where, given an object with a type similar to a -> b, c, and an object with a type similar to c, d -> e, the two objects can be "cut together" to form an object with a type such as a, d -> b, e.
The most obvious is the cut rule in sequent calculus.
Another example is tensors; if the first two types I gave above are the types of two tensors, then we can take the inner product of the tensors to get a tensor of the third type.
A metaphor may be if I have two trees of cables with various plugs and sockets sticking out of them; I can take a plug from one tree and plug it into a socket from the other tree, forming a larger tree.
I don't suppose anyone's written about these sorts of things in general?
aha, so a natural map between functors isn't enough to identify canonically isomorphic objects? or is every natural map between functors, isomorphic to identity?
take for example V and its dual---there's no natural isomorphism, and indeed, even though they are isomorphic, they don't "feel" "same-y"
but for V and its double dual, there is a real sense in which those objects are the same---a real sense in which you can treat vectors as linear functionals on dual-space vectors
that sense being given precisely by the natural iso
@Fargle I see, so (1) A natural transformation between functors whose components are isomorphisms is not what you mean here? (2) "Maps" naturally isomorphic to identity really blur out the distinction between say, $V$ and $V^**$, kinda like $\mathbb{R} \times \mathbb{R}$ and $\mathbb{R}^2$, right?
$\mathbb{C}$ and $\mathbb{R}^2$ are also the same thing (as sets)
but the comparison doesn't really work, because these are just two specific objects
natürally isomorphic functors means that you associate objects to one another and can identify them in ways which respect all the maps arising between them
many definitions of $\Bbb C$ give it as ordered pairs $(a,b)$ of real numbers with $a + ib$ as mere notational shorthand, and in such definitions $\Bbb C$ is literally $\Bbb R^2$
the key thing with double duals is that "double-dualing" is a functor on the category of vector spaces which is naturally isomorphic to the identity functor, and from this, you can for a particular $V$ exhibit a canonical isomorphism $V \leftrightarrow V^{**}$
the number one problem with category theory is that what is really meant by a particular statement is hidden behind, on average, about thirty definitions lol
(anyone who knows better should let me know if I'm off-base here, I woke up like 40 minutes ago and am also stupid)
It might be worthwhile pointing out that saying these functors are naturally isomorphic has a lot more content than just saying that you can exhibit isomorphism $V\cong V^{\ast\ast}$ for each vector space $V$, namely that all these isomorphisms are compatible with the maps between these vector space. Let's make this precise. For each vector space $V$, let $\varphi_V\colon V\rightarrow V^{\ast\ast}$ be the "canonical" isomorphism. What's key to the concept of a functor is that a functor not only associates objects to objects, but also maps between objects to other maps between the correspond…
This is why it makes sense to identify $V$ with $V^{\ast\ast}$. They not only look like one another by themselves (i.e. isomorphic vector spaces), but they and the morphisms out of them look like one another within the context of the entire category.
Maybe look at a different example too. Let $M$ be a smooth manifold of dimension $n$. For each $p\in M$, $T_pM$ is an $n$-dimensional vector space, so it is isomorphic to $\mathbb{R}^n$. This doesn't feel natural: you can't generally tell me an isomorphism without choosing a basis (which comes from choosing a chart) and there is not a meaningful choice of what maps this isomorphism should respect. (There's also a non-categorical reason why these isomorphisms don't feel natural: they don't necessarily come together smoothly if the tangent bundle is non-trivial.)
fwiw, the second case is even nicer, because the isomorphisms there actually come together nicely to give a bundle isomorphism $T(M\times N)\cong TM\times TN$, which is natural in the appropriate setting
Let's see if I can construct a BM by hand. Let $Z_m^n \sim \text{Unif}(1, -1)$ for $1 \leq m \leq n$, $n \in \Bbb N$. Define $X_0^n = 0$, $X_m^n = \frac1n \sum_{k = 1}^n Z_k^n$ for $1 \leq m \leq n$. So $X_m^n$ is basically an $m$-step random walk with step-length $1/n$.
To show how the convolution applied to an image can be viewed as matrix-vector multiplication, let's suppose that we want to apply a $3 \times 3$ kernel to a $4 \times 4$ input, with no padding and with unit stride.
Here's an illustration of this convolutional layer.
Now, let the kernel be de...
