this is really an issue with the codomain, not the domain though
even if $R$ is commutative, you can only evaluate polynomias in $R[X]$ at elements in an $R$-algebra $S$ that commute with $R$
this makes sense: an $R$-module structure on an abelian group $A$ is the same thing as a homomorphism $R\rightarrow\mathrm{End}_{\mathbb{Z}}(A)$ and extending this to an $R[X]$-module structure is the same thing as picking an $R$-linear endomorphism of $A$ as which $X$ acts
@robjohn According to this post, developers are actually looking at the MathJax wrapping bug on the Activity page. The bug has been around since they changed the Activity page over 2 years ago. I still doubt it will be fixed.
Hey all! Random question can the problems and solutions of ethics be regarded as a problem in second order logic? (I'm using second order logic in the Penrose Lucas kinda arguments) ... Apologies if this is too vague
@Jakobian yeah I thought the same. That's a definition of conditioning. I was confused because I misunderstood that $X\in\mathcal{H}$ is an assumption (which is not but rather $E(X\mid\mathcal{H})\in \mathcal{H}$). I can just conditioning on any sub-sigma algebra
@BalarkaSen True but I meant the numbers. There're two numbers I can control for each cusp.
@BalarkaSen Yes I know but it wasn't quite clear to me...
Is there a Wieferich-prime $p>2311$ with respect to base $2024$ , in other words a prime $p>2311$ satisfying $$2024^{p-1}\equiv 1\mod p^2$$ ? I found no more upto $9\cdot 10^9$. The only "Intersting" such prime is $2311$ which happens to be a primorial +/-1.
@onepotatotwopotato It's $(p, q)$, where the Dehn filling slope is $p/q$.
The fact that the two sliders seem to vary over all reals and not just coprime integers I think means that you're looking at some modification of the triangulation in $\Bbb H^3$ which need not necessarily descend to a manifold unless $(p, q)$ are coprime integers.
Hit the button "make manifold" to "smooth it out"
You see the big spherical nodes in the inside view right? Observe that those seem to have some kind of white markings on them which look like lines. If you vary the coefficients the white markings become spirals.
This is because the spherical nodes are the cusp neighborhoods, and the white lines are the slope of the lines getting killed under Dehn surgery.
If it's not rational, you'll get some discontinuous patterns
"Go inside" the cusp neighborhoods by rotating about a cusp to see this more clearly
I am trying to understand the proof of the following fact from Hatcher's Algebraic Topology, section 2.A.
Theorem. Let $X$ be a path connected space.
Then the abelianization of $\pi_1(X, x_0)$ is isomorphic to $H_1(X)$.
I am having trouble understanding the last step of the proof.
Step 1...
@Balarka suppose $X$ is a smooth oriented $n$-cell and $X_1,X_2\subseteq X$ are (closed, codimensions $0$) oriented smooth submanifolds with boundary that are themselves $n$-cells s.t. $X=X_1\cup X_2$ and $X_1\cap X_2$ is a neat codimension $1$ submanifold with boundary of $X$. is $X_1\cap X_2$ automatically an $(n-1)$-cell?
@Thorgott Seems to me that $X_1 \cap X_2 \hookrightarrow X = D^n$ is a properly embedded (i.e., boundary of $X_1 \cap X_2$ goes to boundary of $D^n$) smooth codimension 1 submanifold such that the complementary pieces are smooth codim 0 manifold with corners which are $n$-balls. Doubling this $D^n$ then gives an embedded codimension 1 submanifold $S \subset S^n$ (obtained by doubling $X_1 \cup X_2$) such that $S^n \setminus S$ are $n$-balls. Morse theory forces $S$ to be a topological sphere.
Let $M = X_1 \cup X_2$. This is a smooth manifold with boundary $(M, \partial M)$ such that $M \cup_{\partial M} M = S^{n-1}$. Morse theory should again tell you $M = D^{n-1}$.
Smoothly it's possible something goes wrong. Especially around dimension 4.
It's not known if there are exotic 4-balls.
Ah, wait, one second. It's not true that $M \cup_{\partial M}M = S^n$ implies $M$ is a ball without carefully knowing what is happening near $\partial M$. There are non-ball manifolds $(M, \partial M)$ and a diffeomorphism $f : \partial M \to \partial M$ such that $M \cup_f M$ is a sphere.
@Thorgott Potential counterexample: There is a 4-manifold with boundary $(M, \partial M)$ and an involution $f : \partial M \to \partial M$ such that $M \cup_f M = S^4$ and $f$ extends to a homeomorphism of $M$ (but not a diffeomorphism).
Let $D^5$ be a ball bounding this $S^4$, treat it as a manifold with corners whose boundary faces are the two copies of $M$ and the corner is $\partial M$ where they intersect. Let $X_1, X_2$ be two copies of this 5-ball with corners. Glue $X_1$ and $X_2$ along a pair of boundary face $M$ by the extension of $f$ to $M$. Then this union $X_1 \cup X_2$ seem…
So usually when hyperlinking books etc in recommended books could we use the publishers website instead of the traditional first Google result? Would this create value?
@Thorgott I think it self intersects because the text (image not shared here) says that the image on the right with subspace topology from R^2 is not homeomorphic to R.
so immersed submanifold is used instead of embedded in the definition of Lie subgroup to make the following true: Connected Lie subgroups of a Lie group G are in one to one correspondence with the Lie subalgebras of Lie(G).
I was originally just trying to describe addition in homotopy groups without fixing explicit models for the $n$-cells and their inclusions, but I also didn't need to be as general as in that question lol
hi. so(3) can be thought of as euclidean metric preserving vector fields on $R^3$ and also as the left invariant vector fields on the projective 3-d ball
what is this idea called when a lie algebra can be identified with different vector fields
You take a L. group G. You consider the set K of all left invariant vector fields on G. K with the commutator bracket is called the Lie algebra associated to G. (This is the definition of Lie algebra associated to G).
the other vector field association can be : take $R^3$ and associate an element of so(3) to the vector field which points in the direction of the infinitesimal rotation of $R^3$ generated by that element @Thorgott
The above link I sent is about one-dimensional random walks and expected number of steps before it reaches a point at a distance of a from the origin, given the random walks are symmetric and start at the origin.
Maybe one last question if possible Thorgott...my book says "It is not hard to see that $r \in \Bbb Q \iff \exists (p,q) \in \Bbb Z \times \Bbb N^\times (r = p/q)$
Two questions about this: (1) By $r = p/q$ they presumably mean $r = pq^{-1}$ where $p$ and $q$ are the elements of $\Bbb Q$ that are also in $\Bbb Z$ and $\Bbb N^\times$ respectively, right? i.e. in the subsets of $\Bbb Q$ which are isomorphic thereto?
In that case, I can see the $\Longleftarrow$ immediately by closure of the field $\Bbb Q$
The question is: Determine the expected number of steps of either of the two independent one-dimensional walks $R_1$ or $R_2$ on $\mathbb{Z}$ reaching a point $P$ at a distance of $a$ from the origin. The random walks start at the origin and are symmetric. I would guess that it is $a^2$ and it is because of the independence of the two random walks and this thread: https://math.stackexchange.com/questions/288298/symmetric-random-walk-with-bounds
However, I am not entirely sure whether this result works or not because it would imply that the same is true for more than two random walks which…
@Thorgott Hrm... I don't like that sentence. The definition above gives that $\mathbb{Q}$ is the smallest field containing the integers. The proof is constructive---the authors build a field in the "obvious" way, then show that any other field containing $\mathbb{Z}$ must be at least as a big. But in the construction, a rational number is an equivalence class of pairs of integers.
they define the multiplication in $\mathbb{Q}$ so that the rational number $[(p,q)]$ becomes a fortiori the result of dividing the integer $p$ (interpreted as a rational number in the way they specified) by a non-zero integer $q$ (interpreted as a rational number in the same way again) in $\mathbb{Q}$
But you don't build $\mathbb{Z}$ with an equivalence relation on $\mathbb{N}$. You build it from an equivalence relation on $\mathbb{N}\times\mathbb{N}$.
@Koro no, just by cardinality reasons. But as Xander mentioned you can do equivalence classes of sequences of rationals. Or use subsets of the rationals if you prefer Dedekind cuts
The Dedekind construction gives you the least upper bound property for free, which is nice. And I feel like it also gives you the nested interval property fairly easily (though it's been a while since I've thought about this).
I don't remember the details, but the basic idea is that a real number $r$ is uniquely determined by the function $\mathbb{N}\rightarrow\mathbb{N},\,n\mapsto\lfloor rn\rfloor$, so you try and work backwards from that
So to construct the reals/complete an ordered space, I'll just use Dedekind cuts. To complete a metric space, I'll use Kuratowski's embedding. No need for the Cauchy sequences contruction
And this is better than Cauchy sequences, since you have an isometry of $X$ into $C_b(X)$ with supremum norm (or $C^*(X)$ as I call it)
soumik: often i think there's no "decision" to name something after someone for any specific reason, it just emerges organically out of how people begin referring to a method or result, which can happen for reasons other than priority as often as it happens for that reason. which is very close to your "who popularized it most," except it doesn't require the named-after person to do anything (e.g. all zorn might have done is published some paper)
The relevant notion of "completeness" in metric spaces is Cauchy completeness, i.e. all Cauchy sequences converge in the space. It is natural to attempt to complete a space using this condition. From a pedagogical and historical point of view, it makes sense to talk about the Cauchy construction. Kuratowski's embedding is (like other embedding results) a very nice and useful result, but it is not generally a useful way of thinking about how to construct / complete a metric space.
soumik: if you want something named after you, i think your best bet would be to publish it in a journal that is likely to be read by a lot of people who publish even more than you do :)
@Jakobian I am really tired of you constantly picking fights with people here, and have no desire to engage with you further on this. I am walking away.
If someone has learned real numbers, continuous functions etc., then one has just as much access to Kuratowski's embedding construction as the other one
So its not less useful in pedagogical sense
Moreover, the naturality of Cauchy sequences construction - is it really that natural?
@Koro As I said above, the Dedekind construction has a couple of really nice features---the most "obvious" being the least upper bound property, which is hard to get at otherwise, and is maybe one of the harder-to-understand aspects of the reals from the point of view of a learner.
I can construct a completion using Cauchy sequences in my sleep. It's a tautological process. I can't prove Kuratowski embedding in my sleep. It requires a non-trivial idea.
@Koro If I were teaching a class in which it were necessary to build a model of the reals from some other place, I would want students to know both Dedekind's and Cauchy's approaches, and to be able to show that they are equivalent. I would also want students to be able to articulate the useful things we get from both constructions.
Oh never mind, I think I'm meant to use an earlier corollary which says that if $a \in Z$ and if the equation $x^n = a$ has solutions in $Q$ then $x \in Z$
then recall how you prove \sqrt 3 to be irrational.
How to prove this?
If I apply Ad on both sides, then lhs= Ad(exp(sX))o Ad(exp(tY) because Ad is adjoint representation and that's a homomorphism. not sure how to take it from here.
If $0\to K\to N\to M\to 0$ is a SES of modules, $M$ is finitely presented, $N$ is finitely generated, then $K$ is finitely generated. My interpretation of this result is that if we try to write a finitely presented module $M$ as a quotient $N/K$ where $N$ is finitely generated, then that forces $K$ to be finitely generated as well.
which doesn't sound like a big demand
its a reasonable theorem under that interpretation
I am trying to understand the proof of the following fact from Hatcher's Algebraic Topology, section 2.A.
Theorem. Let $X$ be a path connected space.
Then the abelianization of $\pi_1(X, x_0)$ is isomorphic to $H_1(X)$.
I am having trouble understanding the last step of the proof.
Step 1...
