@LeakyNun Seems like the calculational proof is encoding exactly the same proof the universal properties are. To encode that map, you first swap to $((A \otimes_R B) \otimes_A A) \otimes_R B$, then pair $B \times A \to B$ using the bilinear product. So you use it to collapse $B \otimes_A A \to B$. Similarly with the last term. In every case, the relevant multilinear maps are always A-linear in the first coordinate.
@MikeMiller ok I just did it in lean, and it turns out that the adjunction is sufficient to do the isomorphism without much work
"the adjunction" being $$\operatorname{Hom}_{A \downarrow \textbf{CRing}} (A \otimes_R B, C) \cong \operatorname{Hom}_{R \downarrow \textbf{CRing}} (B, C)$$
If i have a complex function f(z) in terms of z then to check if it is analytic in a domain I just need to see if f'(z) is defined over the dominant right?
To prove it rigorously I need to take the limit definition. But as a check it works if I just differentiate and see if it is defined. Right?
I have to solve the differential equation (x^2+y^2)dy +(2xy)dx=0 where y(0)=1 . Now , I found dy/dx =\frac{-2xy}{x^2+y^2} . This is a homogenous diff eqn. I tried to solve it by substituting y=vx . However, (0,1) certainly does not satisfy y=vx as given in the question. So, is this question missing some details ?
Is there a general rule for when topological spaces of multiple dimensions are homeomorphic to spaces of strictly fewer dimensions, or is it more of a case-by-case matter?
Arbitrarily many dimensions. For just an example $\mathbb{R}$ in standard metric cross $X$ in trivial topology, where $X$ is any nonempty set. $\mathbb{R}$ serves as the first dimension and $X$ serves as a second.
Ah, I don't have a good definition for that. I suppose a topological space being dimensional would require it to be a metric space, to allow distances to be defined?
I was trying to create a quick ad hoc definition from the top of my head and the best thing I could recall was n-dimensional vector fields having a basis of n-many linearly independent vectors.
and I'm not 100% sure that's an accurate recollection.
The whole issue is that you start with asking about topological spaces, but if you use a measure of dimension that only depends on the topology, then clearly no topological space can be homeomorphic to one of smaller dimension
Okay, so a given element of a product space is given by $(x_1 ,x_2 ,x_3 ,...,x_n)$ where each $x_i\in U_i$ for $\{U_i |1\leq i\leq n\in\mathbb{N}\}$ a collection of topological spaces used to create the product space. Since we have $n$ sets in that collection, we say the product space is n-dimensional. Is that any better?
@LeakyNun No idea what happens for monoids in general.
I have only studied the semigroup algebra of monoids in order to determine their special representations in the standard basis, and that was just as an example for a paper
Oh, then what's the established definition? (maybe google will be able to give me that)
Okay, so dimension in topological spaces is ambiguous, as there are multiple things which it could potentially refer to? Such as Lebesgue covering dimension or inductive dimension?
suppose $\sum a_g g \in Z(R[G])$, so $\sum_h a_{h^{-1}g} b_h = \sum_h a_{gh^{-1}} b_h \forall g \forall b$, so $a_{h^{-1}g} = a_{gh^{-1}} \forall g \forall h$, so indeed
@LeakyNun Yeah, this is probably easier to see by writing the algebra as the set of functions with finite support
then being in the center implies taking the same value on all elements of any given conjugacy class (same argument as usual), and then it must be $0$ on any infinite conjugacy class to preserve having finite support.
I can see that the thing in the image also has projective cokernel if thats of any significance (since right exactness preserves cokernel, and S\otimes - has a right adjoint that preserves epimorphisms)
@MatheinBoulomenos Let $\Bbb Z_2 = \langle x \rangle$ act on $\Bbb Z_3 = \langle y \rangle$ nontrivially. Then $H^2(\Bbb Z_2; \Bbb Z_3)$ is $\text{Fix}_{\Bbb Z_2}(\Bbb Z_3)/(1 + x)\Bbb Z_3$, but the numerator is the trivial group (no nontrivial element of $\Bbb Z_3$ is fixed by $\Bbb Z_2$), so $H^2(\Bbb Z_2; \Bbb Z_3) = 0$, no?
I'm confused; there are two extensions of $\Bbb Z_2$ by $\Bbb Z_3$.
Oh, different actions. In the other one the action is trivial.
OK, so this is correct. Says there is a unique extension with that action and it splits.
So the correct, precise, statement here is that given an $H$-module $N$, $H^2(H; N)$ classifies extensions of $H$ by $N$ with that specified monodromy action.
That is correct. $H^n(\Bbb Z_n = \langle x \rangle; A)$ in general is $\ker(1 + x + \cdots + x^{n-1})/(x - 1)A$ in odd dimensions and $\text{Fix}_{\Bbb Z_n}(A)/(1 + x + \cdots + x^{n-1})A$ in even dimensions.
Just a straightforward generalization of the resolution you used
'Cuz $(x - 1)(1 + x + \cdots + x^{n-1}) = 0$
Note how that this means $H^1(\Bbb Z_2; \Bbb Z_3) = 1$, which implies all the complements of $\Bbb Z_3 \leq S_3$ are conjugate; they are all the 2-Sylows, so of course they are conjugate!
$H^1$ can track the conjugacy structure of the Sylows, which is a big thing sometimes.
What are the orbits of the action of a group on it's $p$-Sylows, ie
If I use the extension $\Bbb Z_3 \to S_3 \to \Bbb Z_2$, it seems that the $E^2$ page for $H^*(BS_3)$ consists of $0, \Bbb Z_2, 0, \Bbb Z_2, \cdots$ for rows at multiplies of 3, all $0$'s for 1 mod 3 rows, and $0, \Bbb Z_6, 0, \Bbb Z_6, \cdots$ for 2 mod 3 rows
@AlessandroCodenotti I have difficulty understanding when I'm allowed to do what, I have trouble understanding the difference between predicates, terms, relations, etc. I have trouble understand how variables and constants are interpreted, etc.
@LeakyNun I only know one proof of that: take a generating system of $M'$ send it to $M$ and lift a generating system of $M''$ and take the union of that
this is probably very basic, but how exactly does a SES of topological groups $1 \to G' \to G \to G'' \to 1$ induce a fibration $BG' \to BG \to BG''$? (and do we need some assumptions?)
@LeakyNun here's another proof: one can show that $M$ is f.g. iff for every ascending chain of submodules $(M_i)_{i \in I}$ with $M=\bigcup_{i \in I} M_i$, one has $M=M_i$ for some $i$. Using this, if you have an ascending chain of submodules in $M$ where $0 \to M' \to M \to M'' \to 0$ is exact, you can take the image of that chain in $M''$ and the preimage in $M'$ and then for some $i$, you get that the preimage is $M''$ and the image is $M'$, then for that $i$, $M_i=M$
I doubt proving that characterization and using it is any shorter though
This seems like a stupid doubt but I am going to ask it anyway. If I have an orthonormal basis $X_j$ at a point of a Riemannian manifold $M$ with an affine connection $\nabla$, does it imply that $\nabla_{X_i}X_j = 0$
This comes up because I am trying to prove that if the connection is compatible with the metric iff $\frac{d}{dt}\langle V,W \rangle = \langle \frac{D}{dt}V, W \rangle + \langle V, \frac{D}{dt}W \rangle $ where V,W are vector fields along a differentiable curve $c$ on the manifold
@Albas: The answer is NO, of course it doesn't imply that. That can happen only if the connection has zero curvature. How does it come up in the compatibility proof?
@TedShifrin In the compatibility proof when they show that given compatibility the equation holds, they take an orthonormal basis $P_i$ at a point $c(t_0)$ and since compatibility implies the angles between parallel vector fields remain the same they extend $P_i$ throughout the differentiable curve. Then given any vector field $V$, $V = \sum_{i} v^{i}P_i$ and $\frac{D}{dt}V = \sum {i}\frac {d}{dt}v^{i}P_i$. I don't get how they conclude the last bit.
Shouldn't that be only possible when $\nabla _{P_j}P_i = 0$ because when you evaluate $\frac{D}{dt}V$ there is a nabla term in there as well. I must be missing something obvious
I had never thought through a counterexample on this before. I realized the product of 4-manifolds would be the cheapest place.
@Albas: To emphasize. We can do parallel translation along a curve always. To get a frame that's parallel in all directions requires a flat connection.
hey. what are the stable homotopy groups of an eilenberg-mac lane space K(G,n)? it feels like it should be super simple because, well, the unstable homotopy is simple, but computing the stable homotopy groups involves taking suspensions, and I don't know how those look like!
Hi. Suppose we're given an Ehresmann connection on a fiber bundle $\pi:X\to Y$. Given a curve $\gamma$ in the base $Y$, consider the pullback of the fiber bundle and its connection along the curve. It seems the horizontal bundle of the pulled-back bundle $\gamma^\ast X\to I$ is a line subbundle of the tangent bundle $\mathrm T\gamma^\ast X\to \gamma ^\ast X$. Thus we locally have integral curves in $\gamma^\ast X$ for the horizontal bundle.
It also seems that flowing along these integral curves should give diffeomorphisms between fibers of $\pi$, but I'm not sure how to prove this. Would appreciate some help!
I have a question regarding the idele group $\mathbb{I}_\mathbb{Q}$, which is the restricted direct product of all $\mathbb{Q}_p$ for primes $p$. Given some $y \in \mathbb{I}_\mathbb{Q}$, does there exist a neighborhood $U$ of $y$ such that $gU \cap U = \emptyset$ for all $g \neq 1$ with $g_p \in \{1,-1\}$ for all primes $p$?
short exact sequences are in particular chain complexes, the notion of a chain map is pretty central to homological algebra
that doesn't really answer the question, but I'd say most of homological algebra seems difficult to motivate unless you see it arise from some other context
@MikeMiller ah, right. even hard for G=Z and n=1! Ithink the bottom line is that I had misinterpreted the slogan "if you know the homotopy groups, you know the stable ones" as "if you know all the homotopy groups of a space, then you know all its stable ones, whereas it should be "if you know all the homotopy groups on earth (or just the ones of your space and its suspensions), then you know all the stable ones (or just the one of your space and its suspensions)"
that's what you always do with derived functors in terms of a resolution, you take a resolution $P^\bullet \to X$ of your object $X$, and then you throw $X$ away and consider the cohomology of the complex $F(P^\bullet) \to 0$
Hi. Suppose we're given an Ehresmann connection on a fiber bundle $\pi:X\to Y$. Given a curve $\gamma$ in the base $Y$, consider the pullback of the fiber bundle and its connection along the curve. It seems the horizontal bundle of the pulled-back bundle $\gamma^\ast X\to I$ is a line subbundle of the tangent bundle $\mathrm T\gamma^\ast X\to \gamma ^\ast X$. Thus we locally have integral curves in $\gamma^\ast X$ for the horizontal bundle. These integral curves are also transverse to the fibers of the bundle $\gamma^\ast X\to I$.
