I know $AB$ will undergo an anti-clockwise rotation so $B$ would move a little downwards and left. Can I say $B$ was acted by $F_1$ that caused its motion? ($F_1$ acted only for a moment)
so what happens if you equip G with the indiscrete topology i guess my corresponding Borel sigma algebra only contains {} and G so haar measure would just be the same as assigning a number to G
and in this case it seems like all my measurable functions have to be constant functions?
and in particular you can produce a unit literally given by the constant 1
Then if $P_a = (a_0, \cdots, a_n)$ and similarly with $P_b$, then their convex combination $\lambda P_a + (1-\lambda) P_b$ is given by doing their convex combination termwise.
For us, we start with a k-facet of $\Delta^n$. I assume for convenience it is spanned by $\{x_0, \cdots, x_k\}$. First let's find barycentric coordinates for the barycenter of the facet $\{x_0, \cdots, x_k\} \cup S$, where $S \subset \{x_{k+1}, \cdots, x_n\}.$ We will call this barycenter $P_S$.
Then the barycentric coordinates are $P_S = \frac{1}{|S|+k+1} \vec{a}_S$, where the $j$th term of $\vec{a}_S$ is $1$ if either $j \leq k$ or $j \in S$.
Claim: if $S, S'$ are disjoint, then $P_{S \cup S'}$ is a convex combination of $P_S, P_{S'},$ and $P_\varnothing$.
Oops, that's not true.
Clearly $\vec{a}_{S \cup S'} = \vec{a}_S + \vec{a}_{S'} - \vec{a}_{\varnothing}$. Then we get $$P_{S \cup S'} = \frac{|S|+ k+1}{|S|+|S'|+k+1} P_S + \frac{|S'|+k+1}{|S|+|S'|+k+1} P_{S'} - \frac{k+1}{|S|+|S'|+k+1} P_\varnothing.$$
What are examples of machine learning techniques (i.e. models, algorithms, etc.) inspired (to different extents) by neuroscience?
Particularly, I'm interested in recent developments, say less than 10 years old, that have their basis in neuroscience to some degree.
Write $P_i = P_{\{i\}\}$, where $i > k$. Then the above shows that any $P_S$ lies in the hyperplane spanned by $P_\varnothing and the $P_i$, which is an $(n-k+1)$-dimensional hyperplane.
This hyerplane is also transverse to the hyperplane the simplex lies in (the one where the coordinates sum to 1).
So the intersection of this hyperplane with the simplex --- which the $P_S$ all live in --- lies in an affine hyperplane of dimension n-k.
One checks that you cannot span one of the $P_i$ with $P_\varnothing$ and the other $P_j$ (which is clear), and thus they do not lie in a smaller-dimensional hyperplane.
So you lie in a convex set, which spans an affine hyperplane of dimension (n-k). Thus the convex set is (n-k)-dimensional, and thus equivalent to a ball.
Ok, assume $f\colon\omega_1\rightarrow\mathbb{R}$ is continuous. Let $M=\sup f(\omega_1)\in\mathbb{R}\cup\{\infty\}$. Then there exists a sequence $(\alpha_n)_n\in\omega_1^{\mathbb{N}}$ such that $f(\alpha_n)\nearrow x$ strictly. Now, using a peak lemma type argument, there should be a subsequence $(\alpha_{n_k})_k$ that is strictly monotonic. Strictly monotonically decreasing would violate regularity, so it must be strictly monotonically increasing. Then $\alpha:=\sup_k\alpha_{n_k}=\bigcup_k\alpha_{n_k}$ is a countable limit ordinal (countable as countable union of countable sets and a lim…
Hello! I have a question regarding invertible functions. Say I have a function $f:\subset\mathbb{R}to\mathbb{R]$. If I find an inverse function $f^{-1}$ using some computations, and then I find the jacobian matrix of both $J_{f}$ and $J_{f^{-1}}$.
Actually, I don't need a monotonic subsequence. All I need is a subsequence without a maximal element. This surely exists: If $(\alpha_n)_n$ has a maximal element, remove it, if the resulting sequence has a maximal element, remove it, etc.. This process must terminate in finitely many steps for otherwise we get an infinite decreasing sequence of ordinals, which violates regularity, so after those finitely many steps, the resulting subsequence has no maximal element.
1) Say an affine combination of vectors means the coefficients sum to 1; affine combinations of (n+1) affinely independent vectors span an affine n-dim hyperplane.
And the last fact should really not be too unclear: you average out the corresponding barycenters of the (k+1)-facets, and then need to subtract off a multiple of the barycenter of the k-facet due to overcounting.
The last thing you need is: let X be the convex hull of some number of elements in R^n, so that those elements span R^n. Then X is homeomorphic to a ball.
This is probably obvious but I only see it via an inductive argument.
@MikeMiller Suppose $S = \{x_{i_1}, \dots, x_{i_l}\}$, then is $P_S$ an affine combination of $P_\varnothing$ and $P_{i_1}, \dots, P_{i_l}$? I don't see how you concluded that the dimension of $P_S$ is (n-k+1).
@feynhat $P_S$ is a point, so its dimension is 0. And yes, that's what I said above.
(yes it is such an affine combination).
The claim is that the affine space spanned by all of the $P_S$ is indeed spanned just by the ones from the k-facet and the (k+1)-facets, which are independent, which gives the dimension count.
I will re-write the argument as cleanly as possible, leaving a couple steps for you as exercises.
I'm reading something else so will progressively be less available.
1) Forget the previous notation, it was confusing. Let $S$ be a subset of $\{0, \cdots, n\}$, and hence an $(|S|-1)$-simplex. Write $B_S$ for its barycenter.
2) Claim: Given any two subsets $S, S'$, $B_{S \cup S'}$ is an *affine* combination of $B_S, B_{S'}, B_{S \cap S'}$. Write this in terms of the formula for the barycenter, $$B_S = \sum_{i \in S} \frac{x_i}{|S|}.$$
3) Corollary (induction on $|S|-k$). For any set $S \subset \{0, \cdots, n\}$ containing a fixed $(k+1)$-element subset $\sigma$, we may write $B_S$ as an affine combination of $B_{\sigma}$ and the $n-k$ points $B_{\sigma \cup \{i\}}$ for $i \not\in \sigma$.
(Maybe changing notation after you've mastered the previous notation is foolhardy, but oh well.)
4) Corollary: the set of all barycenters of sets $S$ containing $\sigma$ lie in the affine hyperplane spanned by $B_\sigma, B_{\sigma \cup \{i\}}.$
5) Claim: These $n-k+1$ points are affinely independent. (Exercise, follows from the $x_i$ being affinely independent.)
6) Claim: Any $m+1$ affinely independent points span an $m$-dimensional affine hyperplane.
7) Corollary: the collection of $B_S$, where $S$ varies over all sets containing $\sigma$, span an $(n-k)$-dimensional affine hyperplane.
8) Claim: Let $y_i$ be any finite collection of points in $\Bbb R^m$, which affinely span $\Bbb R^m$. Their convex hull is $m$-dimensional. (Proof: it has interior.)
9) Any convex hull of finitely many points is homeomorphic to a ball of the appropriate dimension. (This is a standard fact in polytope theory, but I don't know a reference or the proof off-hand, though I have one in mind.)
This gives you the desired claim.
I think all of these claims except (9) are relatively straightforward lemmas.
if $C$ is your convex hull, you get a map $F: S(T_0 \Bbb R^n) \to \partial C$ by sending each unit vector $v$ to the unique $p \in \partial C$ with $p = tv$ for some $t > 0$
It is straightforward that this map is a bijection between CHaus spaces so
Then your map $\tilde F: D(T_0 \Bbb R^n) \to C$ is given by $\tilde F(tv) = t\tilde F(v)$
Here's a class of counterexamples for the pointed homotopy category of connected CW complexes (so even this restriction does not save you). Let $hCW_{\ast}$ denote this category, and let $\pi_{\bullet} : hCW_{\ast} \to \text{Set}_{\bullet}$ denote the functor taking a pointed CW complex to its ho...
Is there a condition weaker than metrizability that ensures that a topological space $X$ for which every continuous $f\colon X\rightarrow\mathbb{R}$ attains a maximum is compact? Or, even better, is there a classification of spaces $X$ with this property?
I mean when you glue spaces by a pushout large amount of information gets collapsed. Take some large space and map it into a point, and some other random shit happening in the background
The point of ho(co)lim is that they fatten the spaces appropriately so that these operations are remembered
but i mean examples where this isn't true come very quickly you don't need to be fancy. let the base be D^2, and take the pullback of the inclusion of 0 and the inclusion of the boundary circle
clearly not the same as when you crush D^2 to a point
@BalarkaSen I was thinking about noncompact examples (there are some which are pseudocompact but neither compact not sequentially compact in Steen and Seebach of course, I just checked)
Assume $f\colon X\rightarrow\mathbb{R}$ is bounded, but doesn't attain it's maximum. Let $M=\sup f(X)\in\mathbb{R}$ and fix a sequence $(x_n)_n$ in $X$ such that $f(x_n)\nearrow M$. We have $f(x)\subseteq(-\infty,M)$. Fix a homeomorphism $g\colon (-\infty,M)\rightarrow\mathbb{R}$ such that $g(x)\rightarrow\infty$ as $x\rightarrow M$ (i.e. the standard such map) and consider $g\circ f$. Then $(g\circ f)(x_n)\rightarrow\infty$ as $n\rightarrow\infty$.
So if all continuous real-valued maps are bounded, they all attain a maximum. Or am I talking nonsense?
oh right, now that I think about it, all I did in my earlier proof is to show that $\omega_1$ is sequentially compact and then pull through with the standard proof
@Thorgott To rephrase: we consider the collection of subsets of $\Bbb R$ of the form $f(X)$. Suppose we are further given that every $f(X)$ is bounded; then $f(X)$ is closed. For if $y \not \in f(X)$, then post-compose with the map $g: \Bbb R \setminus y \to\Bbb R$ given by $x \mapsto 1/(y-x)$. Because $gf$ is bounded, we see that $f(X)$ does not accumulate towards $y$. Therefore if every real function is bounded, they also all have closed image.
@feynhat No, I mean dimension as a topological space. We see in (9) that this convex hull is a ball, and we need to know its dimension.
A 2-dimensional convex hull need not be the convex hull of 3 points, or every 2D convex hull [of a finite set] would be a triangle. But of course, convex n-gons exist (convex hulls of n+1 points).
Yes, the hyperplane in (7) is equal to the hyperplane in (4). (This is what (3) gives us.)
Yeah, I understand. Poincare's original proof can actually be made rigorous though, which is what I meant I thought you were reading. I've never actually read through Poincare's argument itself, just other people talking about it (until the picture was clear).
The one thing I can't actually remember is how to construct the actual dual submanifold.
@BalarkaSen He did, the point is that you have an intersection pairing between the given triangulation and the dual cell complex which is nondegenerate by the way you've built it
I feel like I remember there being a formula for which cells to add to get the dual guy though
Right, agreed. I remember being mildly surprised when I saw the intersection thing built in in the definition of cup product as well (the pair of faces you evaluate the two cochains on intersect in a lower dimension face)
(For instance observe that to get the Poincare dual to a vertex in the hollow tetrahedron = S^2, you have to sum over all the dual cells, not just the dual cell to that vertex, which very much is not a cycle)
Probably the smart play is to do it for RP^2.
I don't have time to do that computation today though
Hi chat, I don't wanna interrupt you, but if any of you know by chance a bit of toric geometry, I'd be very happy to discuss about this question regarding an isomoprhism of the group of $T$-invariant Cartier divisors
I had to pick between AG and class field theory in the first semester of my masters (could only do one due to overlapping classes times), I did AG because it seemed like an important topic to know something about, but I think I would have enjoyed class field theory more
I think classical AG with a strong emphasis on historical development of the subject would be beneficial. It's one of the most historically important subjects along with number theory, but most syllabi I have seen seem to fall short of this expectation.
@Thorgott Even though I never felt like I understood any AG properly, I still did quite a bit of stuff that relied on stuff that only even makes sense with schemes
@BalarkaSen Sure. I am not looking for anything precise. I just wanted some geometric intuition for it. The diagram chasey proof doesn't really tell much.
What are examples of machine learning techniques (i.e. models, algorithms, etc.) inspired (to different extents) by neuroscience?
Particularly, I'm interested in recent developments, say less than 10 years old, that have their basis in neuroscience to some degree.
