Today's thought: Analysis - The study of math Geometry - The study of shape Algebra - The study of properties Topology - The study of combination Logic - The study of questions Category Theory - The study of studies
A question can never be answered if it isn't asked. (Note: I might not answer, I'm off doing other stuff. But someone else might swing around and see the question)
@Astyx thanks for your answer, I forgot to add some informations sorry about that... My definition of a projection matrix is the matrix of a projection on a certain basis
@LeakyNun btw, in problem sheet 2, 4b. of algebra 3, do you think there is a typo? is it supposed to say compute $gcd(2+4i,13+ii)$ in $\mathbb{Z}[i]$ ? Since $13+i$ isn't in $\mathbb{Z}[\frac{-1 + \sqrt(-3)}{2}]$ as far as I can tell
@love_sodam you need these lemmas: 1. if K/E is separable and E/F is separable then K/F is separable 2. construction of splitting field 3. if E/F and p in F[x] separable then p in E[x] separable 4. factor of separable polynomial is separable
@love_sodam If f is a separable polynomial over F, let a be a root of f, and E = F(a). Then any embedding F -> barF has deg(f) = [F(a) : F] many extensions E -> barF (adjoin any other root of f, you get an isomorphic field E, embed via that).
Keep adjoining roots, use multiplicativity of degree and "number of embeddings E -> barF sitting over a fixed embedding F -> barF" to conclude
Well, I guess you're not fixing a universal field on the codomain in your definition of separability. Anyway, these are all equivalent: for any finite extension E/F, |Emb_i(E, barF)| <= [E : F] where i : F -> barF is a fixed embedding and Emb_i(E, barF) is all embeddings sitting above i, with equality iff E/F is separable.
@Thorgott If f\in F[x] is a separable polynomial and f = f_1\cdots f_n be a factorization then, if we let f_i(a_i) = 0 where a_i is an element of splitting field of f_i then the splitting field of f would be E = F(a_1,...,a_n) and once we prove F(a)/F case then F(a_1,...,a_n)/F(a_1,...,a_{n-1}) inductively?
@LeakyNun @Thorgott Let $E/\Bbb C(z)$ be an algebraic extension obtained from adjoining a root of $w^n + f_{n-1}(z) w^{n-1} + \cdots + f_1(z) w + f_0(z) = 0$. For a generic choice of $z \in \Bbb C$, this polynomial in $w$ has $n$ solutions over $\Bbb C$. Assume whenever $z \in \Bbb C$ such that this polynomial has less than $n$ solutions, it has exactly one solution over $\Bbb C$.
@Thorgott So what I should do is adjoin all roots of f one by one and every time I do that, F(a_1,...,a_n)/F(a_1,...,a_{n-1}) is a separable extension as each $f_i$ is separable right?
This is the algebraic translation of having a branched cover $X \to \Bbb{CP}^1$ of degree $n$ such that at each branch point the cover looks like $z \mapsto z^n$.
Then it's automatically regular
Pure topology proof: It suffices to prove $\pi_1(\Bbb{CP}^1 \setminus D)$ acts transitively on the fibers, where $D$ is the branching locus. It suffices to take a fiber very close to one of the branch points, so the situation is like the fiber over $1$ under $\Bbb C \to \Bbb C$, $z \mapsto z^n$. Then under monodromy around a loop in $\Bbb C^\times$ any point on the fiber is taken to any other point because it's a cyclic monodromy.
If $(\pi_g,V)$ and $(\pi_h,W)$ are representations of two Lie algebras $\mathfrak g$ and $\mathfrak h$ respectively, what is the representation $V\oplus W$ of? It can't be $\mathfrak g\oplus \mathfrak h$ since this is the representation on $V\otimes W$, is it just $\mathfrak g\times \mathfrak h$ or something?
set-valued functors on a category admitting coproducts are representable iff they have a left adjoint
adjoint functor theorem says that on a complete category, a functor has a left adjoint iff it preserves small limits and satisfies the solution set condition
@123 How exactly the inner product is executed depends on the inner product space you are considering, the only requirement is that it satisfy the axioms of an inner product, other than that you can basically go crazy with it
like for all objects in the codomain, you can each find a set worth of objects in the domain such that all morphisms from that object in the codomain factor through the images of morphisms under those set many objects or sth
As I said, exactly how one executes the inner product depends on the space you are considering, the only constraint is the axioms of the inner product. If you want to know why exactly mathematicians in the past decided on those particular axioms then I can't give you the answer, you'll have to ask a mathematical historian I guess.
A definition is true by definition of what a definition is. The choice to make such a definition is a historical question, to which I don't know the answer, try asking here.
just have to remind myself of how restriction/extension adjunction worked
ah, it should be $R\otimes_{\mathbb{Z}}R$
a group hom $f\colon R\rightarrow M$ gives the $R$-hom sending $r\otimes r^{\prime}$ to $r^{\prime}f(r)$ and an $R$-hom $\varphi\colon R\otimes_{\mathbb{Z}}R\rightarrow M$ gives a group hom $R\rightarrow M$ by post-composing with the map $R\rightarrow R\otimes_{\mathbb{Z}}R$
Let $X$ be path-connected. What conditions do I need to impose on the space $Y$ such that $f\colon X\rightarrow Y$ is continuous iff its graph is path-connected as a subsapce of $X\times Y$? (the forward direction always holds, of course)
Argument goes like this. Assume $x_n\rightarrow x$. This sequence lies in some interval $[a,b]$. Pick a path $\varphi\colon[0,1]\rightarrow\Gamma_f$ from $(a,f(a))$ to $(b,f(b))$. By IVT, there are $t_n\in[0,1]$ s.t. $\varphi(t_n)=(x_n,f(x_n))$. $[0,1]$ is compact, so pick a convergent subsequence $t_{n_k}\rightarrow t$, implying $f(x_{n_k})\rightarrow f(x)$ and that's actually enough to imply continuity.
This standard argument doesn't generalize at all, tho
For $\Bbb R^2\to\Bbb R^2$ there are easy counterexamples unless I'm missing something? Cut $\Bbb R^2$ along half of the $x$ axis and move the two flaps up and down
@LeakyNun I think to make my argument algebraic you need something like this: If $E/\Bbb C(z)$ is a finite extension such that the corresponding Riemann surface has a branch point at $z=\alpha$, then complete the guy with respect to the valuation of order of functions vanishing at $\alpha$
Then you get a field extension $E_\alpha/\Bbb C(z)_\alpha$ which should keep track of the branching structure at $\alpha$
I don't know enough about valuations to know if that's how it works
If I had a function $h$ defined as: $$ h(n,0) := n \\ h(n,m+1) := f(h(n,m)) $$ where $f:\mathbb{N} \rightarrow \mathbb{N}$ and $n,m$ are natural numbers. And suppose everything is well defined as well as function $h$. My question is: Can I say that for some fixed $n$ , $h(n,n) = f(f(....f(n)...)$ (where number of $f$ calls is equal to $n$) without proving it with mathematical induction ?
Suppose I got this: $$ h(x,z,0) = x \\ h(x,z,y+1) = H(h(x,z,y), z-y)$$ And this as well: $$ F(x,0) = G(x) \\ F(x,z+1) = F(H(x,z),z))$$. I want to prove that $F(x,y) = G(h(x,y,y))$ (if it's correct by all means) using mathematical induction for a fixed $x$. And somewhere in the process I've reached the point that if I use the "unwrap" method I can prove it. But using purely induction, I'm stuck.
Now if I'm trying to prove something that's wrong ,this gonna be pretty embarrassing.
For the regression model $Y = \beta_o + \beta_1x_i + \epsilon_i$. Why is the estimator line $\hat{Y} = b_o + b_1 x_i$ and not $\hat{Y} = b_o + b_1 x_i + \hat{\epsilon_i}$
To a physicist the "generators" of a Lie algebra are just a basis for the vector space structure of the algebra, apparently this means something different to mathematicians, does anyone know what this is?
presumably one would say some collection g_1, ..., g_n are generators for the Lie algebra if the smallest subalgebra containing them is the lie algebra itself
aka, the lie algebra is the span of all of the g_i, together with all [g_i, g_j], together with all [g_i, [g_j, g_k]], etc
@MikeMiller I think it's a useful concept, given a Lie algebra g and a Lie subalgebra h, you can ask what's the least n such that g = h + [h, h] + [h, [h, h]] + ... (n terms here)
That's the bracket-length, I think relevant in sub-Riemannian geometry
If the polynomial $x^3+3px+q$ has a factor of the form $(x-a)^2$, then prove that $q^2+4p^3=0$.
