So your answer is wrong for the tree T_n with one central vertex and n leaves. Then the subgraphs of this are either T_k for smaller k or the discrete graph with n nodes. The average degree of T_k is 2k/(k+1) while the average degree in the discrete graph is 1. So in this specific case I get that your notion seems to be 2n/(n+1) < 2.
Let me reconstruct your proof. It suffices to show average degree is at most 2 for any forest. But sum of degrees is equal to twice the number of edges, which is at most V - 1 --- less if this forest has more than one tree. So we top out at 2(V-1)/V.
@BigSocks Your categorical language is obscuring, not revealing. Yes, there is a set of rational maps between two varieties, no, you cannot take the disjoint union of P(N) varieties under any reasonable notion I know.
@TedShifrin I haven't done anything yet. But I will mention the inductive definition with cofactors. To me this is just an exercise about how swaps affect sign.
You are right @MikeMiller that it is obscuring, but it is sometimes efficient. I don’t see how the second bit implies it does not have small coproducts. I guess continuum is a set and disjoint union is coproduct?
@BigSocks If something is efficient but obscuring it is garbage, throw it out. The whole point of mathematics is understanding and to move in the opposite direction is bad math!
But, yes, the rest is correct.
To say the phrase coproduct of varieties but not understand the picture of coproduct of varieties breaks my heart.
I suppose @BigSocks it depends what a variety is to you. Are they quasiprojective? Then very much NO (projective space has a cardinality bound). Are they locally quasiprojective? then ok.
@Mike Take a complex matrix. For each row, sum the absolute values of its elements except for the diagonal element. Take the disk that has this sum as radius and the diagonal element in the row as center. The union of these disks (they're called Gershgorin disks) contains all the Eigenvalues of the matrix.
@BigSocks Uh all of these are projective varieties, the stuff I was talking about is general nonsense closer to schemes. What you're talking about all takes place as closed subsets of CP^n.
You don't even have infinite coproducts much less coproducts indexed on the reals.
If it helps your intuition you are asking "Is there a category of closed smooth manifolds which is locally small and has small coproducts?" which translates to: "Is C^infty(M, N) a set, and can I take arbitrary disjoint unions of closed smooth manifolds and get a closed smooth manifold?" Certainly not, infinite disjoint unions are not compact
Then you say 'ok now general manifolds' and it depends on definitions.... is a manifold a subset of R^n? Then no, you can only fit countable disjoint unions in R^n. Is it a paracompact Hausdorff etc etc? then fine...
Well, you need to wait for eigenvalues/eigenvectors. I assume you do complex matrices? I never did in my intro courses (just too many things to fit in).
cool addition (this isn't just linear algebra anymore though): if you can partition the $n$ Gershgorin disks into a set of $k$ and a set of $n-k$ Gershgorin disks such that the unions of these respective collections are disjoint, then the former contains exactly $k$ and the latter exactly $n-k$ Eigenvalues (counted with multiplicity, of course)
@Ted Depends, I feel the instructor might do more of linear algebra than group theory because we do not have a proper course on linear algebra (these damn algebraists). If they plan to do that then maybe complex matrices can come in
I do my best but I don't think a course to my standards is possible with standard time constraints and standard expectations of the students' out of class time.
I was lucky. I had a linear algebra/ODE course freshman year. And then I relearned linear algebra in Artin's algebra course sophomore year. But, as always, much of the stuff in these courses I didn't really learn until I taught it and then wrote the books.
I only understood cramer's rule yesterday when I was trying to prove it for myself. I thought about what the goal was and it was completely obvious what to do.
But I have no idea how to communicate it to someone who doesn't think "find a good basis for the dual space" is an obviously good idea.
The point being that letting one of the columns of your matrix vary gives you a basis psi_1, ..., psi_n of row space so that psi_i(v_j) = det(A) delta_{ij}. Now calculate what these row vectors are and there is a formula for A^{-1}.
@Thorgott Formula for A^{-1} in terms of determinants of minors.
There are other ways to describe it eg cofactor expansion to solve Ax = b. They are the same idea so whatever.
You have only described a different phrasing of the same proof. I am describing a proof of the formula for A^{-1}. Your argument is the exact same idea, but applied to solve a different question.
The usual proof I've seen in linear algebra books uses the classical adjoint formula for the inverse to derive Cramer. I'm just saying that's a horrid proof.
We agree that there is exactly one fundamental idea, though: duality between V and Lambda^{n-1} V, and this is expressed by determinant. You are saying: "Solve Ax = b using duality." I am saying: "Find A^{-1} using duality." If you then want to solve Ax = b you can, but this adds an extra step.
I agree that if *your only interest* is in solving the first question then you have added an unnecessary step. However, the first step which we both do is the same step.
This is what I mean when I say they are the same fact: it is an identical idea which goes into the proofs of the two facts.
The fact that you can use my result to obtain your result obscures that your result is of intrinsict interest and can be done with more computational ease, sure.
I also like this problem from the textbook (which I will not use): Find a noninvertible 2x2 matrix whose entries are all distinct prime numbers or prove that this is impossible.
@Astyx ?????????????? $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is noninvertible (over R, say) iff $ad - bc = 0$. By assumption a,d,b,c are distinct primes. By uniqueness of prime factorization we have $ad \neq bc$.
But as followup to the above: there is a nice one on Vandermonde determinants which I think will be difficult for them, and one on the (n-1)ary cross product in R^n
Screw it, I'm including them, they have two weeks since there's a break coming up. I will just warn them that they are more difficult and they should get in touch if they have questions.
I will not include any pointers about how difficult a problem is and instead just include the generality that I will be glad to help in general even though there will be a break.
Thanks for that tip and the tip on Cramer. :) They are both useful (even though I like to argue).
Well, so you can convert so solving a congruence mod 7 and solving a congruence mod 17, which are much easier. What's $40^{478}\pmod 7$ and $40^{478}\pmod{17}$?
What metric on the upper half-plane? I am going to take a metric of the form $dx\otimes dx + f(x)dy\otimes dy$ for some nonconstant (and positive) $f$.
@AlessandroCodenotti idk you could define the product type $\Pi(\Delta[2],A)$ for some type $A$ and then this is formally a homotopy between two elements of $\Pi(\Delta[1],A)$, which are witnesses of equality
whether that actually means anything is another question
He's worried about issues about proper classes (for some inscrutable reason). I think he might think that a set that's bigger than countable is a proper class. I'm having a hard time telling what he thinks. Also, I'm talking about him in the third person while he's right there, which is probably uncomfortable.
@user2103480 I mean there’s gotta be enough structure to get a subvariety. I am talking the alg. geo. sense, not universal algebra sense, of variety fwiw
Everyone is talking past each other because nobody recognizes the points that are actually important here. When you say "The collection of subvarieties of a variety" then that's a set. That's also obviously a set. A subvariety is a subset of V satisfying some conditions... so an element of the power set P(V). P(V) is a set. A subcollection of a set is a set. Done.
@MikeMiller i put it in italics because it was not made clear to me so that is part of what I have to answer for myself. I was guessing maybe separator as in category theory, but you need small colimits and we went through that. If we just had a proper class of iso. classes for some object in the category of varieties I could dispel the notion of “separators” being an adequate description of “test-object” for me
@TedShifrin so i basically know how it works, since i use it as well, you have 10 ticks between each number and the other and it just goes exponentially ... but sometimes lets say between number 3 and 4 you have only 4 ticks.. what does these four ticks represent. i will upload a small picture
@MikeMiller yesn't. I totally get what you mean, any usual operation gives us a set. But if you consider things that embed into a thing, that's often a class
Anyhow, you're splitting the interval from 3 to 4 into 5 equal pieces. Splitting the log into 5 equal pieces means you're multiplying by $(4/3)^{1/5}$ and then $(4/3)^{2/5}$, etc.
@MikeMiller this is what I will write on a scrap of paper and hide under my pillow for when the set theory nightmares hit and I wake up in a cold sweat
Ok, my previous answer wasn't correct. Now, everything works.
First of all, I thank Sergio Buschi, who provided me a counter-example to the OP. Since he gave me his permission, I'll write the details. Let $\mathcal C$ be the subcategory of $\mathbf{Set}$ whose objects are all the sets. As morphi...
In various categories one encounters what are referred to as 'test-objects'. For example, singleton set serves as test-object (in testing the equality of functions) in the category of sets. In the category of graphs, we need two test objects: dot D and arrow A to separate graph maps (Conceptual...
In various categories one encounters what are referred to as 'test-objects'. For example, singleton set serves as test-object (in testing the equality of functions) in the category of sets. In the category of graphs, we need two test objects: dot D and arrow A to separate graph maps (Conceptual...
