One of my profs said to a student who was asking about a paper of Witten "I'll look into it and try to understand it. If I'm stuck, I'm going to give Edward a call"
so an exact sequence $A \xrightarrow f B \xrightarrow g C$ can be factored to $0 \to \ker f \to A \to \operatorname{im} f \xrightarrow {\operatorname{id}} \ker g \to B \to \operatorname{coker} f \xrightarrow \sim \operatorname{im} g \to C \to 0$, three exact sequences connected by $\operatorname{id}$ and $\sim$ @MatheinBoulomenos
@TedShifrin Q: given an odd number $n$, and $P \in \mathcal L(\Bbb R^n)$, and $PP^T = I$, and $\det P = 1$, prove that $\det(P-I) = 0$. Can I do it without any complex numbers?
@TedShifrin what are the possible complex eigenvalues though? $\langle Px, Px \rangle = \langle x, x \rangle = \lambda \overline{\lambda} \langle x, x \rangle$, so anything in $\Bbb S^1$ will do?
@LeakyNun any $P\in{\rm SO}(n)$ is similar to a block diagonal matrix whose blocks are 2x2 rotation matrices; if $\theta$ is the angle of one of these rotation matrices, then $e^{i\theta}$ and $e^{-\theta}$ are complex eigenvalues of $P$ (and conversely)
@Daminark the grad student who graded the assignment actually wrote "! Undefined control sequence" on the paper and said to him that he needs to use dollar signs next time
complexifying isn't just a sneaky algebraic trick; the whole point of complex numbers is to encapsulate rotation, and hey, we're talking about rotations
Suppose we have the trivial bundle we want to prove that it has global section that provide a basis for each of the fibers iff the vector bundle is trivial.
@MatheinBoulomenos great, I'll be sure to write that in tomorrow's test
$(\lambda_2 - \lambda_1) \langle v_1, v_2 \rangle = 0$, but since $\lambda_1 \ne \lambda_2$ and the action of $\Bbb R$ to $\Bbb R^n$ is faithful, we have $\langle v_1, v_2 \rangle = 0$
@Daminark Given an element of e we can write it as $e = \alpha_1 \sigma_1(x_{e}) + \alpha_2 \sigma_2(x_{e}) + \ldots + \alpha_n \sigma(x_{e})$. Ofcourse here under the assumption that $E_x = \{x\} \times R^m$. We assume that our VB is of rank m. so we have the following map $E \rightarrow X\times R^m$ as follows $e \mapsto (x_{e},\alpha_1 + \ldots + \alpha_n)$ this map is continous and has continous inverse as well.
Given any object C (represented by the line), if the second isomorphism theorem holds, then it means lines A and B are of the same length (i.e. A and B are isomorphic)
$(\lambda_2 - \lambda_1) \langle v_1, v_2 \rangle = 0$, but since $\lambda_1 \ne \lambda_2$ and the action of $\Bbb R$ to $\Bbb R^n$ is faithful, we have $\langle v_1, v_2 \rangle = 0$
The middle line rule is always valid because inverting the zero object is one of the most pathological mathematical objects in existence and no mortals will dare to touch it, therefore for all normal applications, the middle line will never had a chance to expand into a triangle again
Or in maths speak, only for a relation of the form $q : 0 \mapsto A$ where $|A|\neq 0$ can invert the zero object
The above 6 examples illustrates the following: 1. Arbitrary o surjective = surjective 2. Arbitrary o injective = injective 3. Surjective o injective = bijective 4. Injective o surjective = bijective 5. Relation o non total function = non total function 6. Relation o function = function
