@TedShifrin Got a question for you about determinants as volumes when you have a moment. I'm trying to see how the linearity of the determinant w/r/t each of its columns goes through. (the scalar multiplication bit is obvious, but I'm not grokking the additivity bit)
(maybe I should just dwell in contemplation of a parallelogram for a bit)
There are some real artists in that Calc 3 course I'm grading, though. Like, they drew a nearly perfect representation of a hemisphere extending from the xy plane into the negative z, with shading and everything. It looked like a bowl.
I had a student years ago at MIT who'd been an architect. He was the official note-taker for the class of 350 students. His pictures — done in India Ink in pen — were stunning.
I actually posted one of his extra credit problem solution on the web many years ago: A picture of the intersection of three orthogonal cylinders. He did a blow-up diagram like you'd get to assemble something from Sears (if anyone remembers them).
I suppose an architect needs to be able to accurately represent what sort of structures they are envisioning. Makes me think about Escher, though, and how he managed to capture mathematical concepts in art form without really knowing the math.
He actually wrote some lectures that were supposed to be given in Canada, but he fell ill and didn't give them. They were published, nevertheless, and this was the beginning of my seminar.
Take a parallelipiped of three vectors u,v,w. Let P be the plane through the origin perpendicular to u. Project the parallelepiped onto P to get a parallelogram
Could someone clarify this hint for me? As it'd probably be useful, know that $U=\text{Span}(u_1,u_2,u_3)$ and $V=\text{Span}(v_1,v_2)$. Anyways: (Hint: Form the matrix $M$ whose rows consists of the following vectors: $$(u_1,0),\quad (u_2,0),\quad (u_3,0),\quad (v_1,v_1),\quad (v_2,v_2)$$ all with length $10$. Then reduce $M$ to a reduced row Echelon form where the row vectors have the form $(w_1,w_2)$ and $(0,w_3)$ where each $w_i$ has length $5$. Explain why all the nonzero $w_1$ form a basis for $U+V$ and all the nonzero $w_3$ form a basis for $U\bigcap V$.)
$$
\text { A group has } 20 \text { Actuaries and } 11 \text { Biostaticians in it. All } 31 \text { people look different (no twins!). }
$$
a. $$
\begin{array}{l}{\text { I asked everyone in the group to send me an email. As it happened, ALL the Actuaries }} \\ {\text { responded before ANY of ...
Consider the function $t(x)$ defined as :
$$ x_1 = x $$
$$x_2 = x $$
$$ x_3 = 2 x^2 $$
$$ x_4 = 4 x^4 + 2 x^2 $$
and for $n > 4 $
$$ x_{n} = \frac { x_{n-1}^2 + x_{n-2}^2 + x_{n-3}^2}{x_{n-2} + x_{n-3} + x_{n-4} } $$
If the sequence converges to a constant then we define $t(x) = \lim x_n $.
...
@LeakyNun I don't think the topology of Spec Z or the structure sheaf play any role in this
not that you can replace "field" by integral domain in the statement
and you can replace Z by any commutative ring A and then you can look at the categories of integral domains B which are A-algebras such that the kernel of A->B is some fixed prime ideal
or you can also do the same thing with categories of fields instead of integral domains
I tried to check Wikipedia for people from Netherlands with surname Kuipers. The Wikipedia articles Nick Kuipers, André Kuipers and Björn Kuipers list this as pronunciation: [ˈkœypərs]. (One of them even has an audio.)
Hello. I have a quick functional analysis question. Let $A=\frac{1}{2}(T+T^\ast )$ be the self adjoint part of $T$. Is something like $\|A^2-A\|\leq \|T^2-T\|$ true?
So, there is a result in our matrix analysis book, that if $\sum_{i=1}^p f_i=id_E$ where each $f_i:E\rightarrow E$ is a linear map (and there are $p$ of them), and $E$ is a vector space over $K$ then the following are equivalent: $f_i^2=f_i$ for all $i$. $f_i\circ f_i=0$ for all $i\neq j$.
However, in the case $E=\mathbb{Z}_2^3$ then the three matrices: $f_1=\begin{pmatrix}1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 0 \cr\end{pmatrix}$ $f_2=\begin{pmatrix}0 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 0 \cr\end{pmatrix}$ $f_3=\begin{pmatrix}0 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr\end{pmatrix}$ are a counterexample to this, correct?
example exercises in our LA courses: prove that $\Bbb Z[i]$ is an Euclidean domain Show that a ring is Noetherian (ACC on ideals) iff every ideal is finitely generated etc.
we did the whole proof of STFGMPID
@Rithaniel take the example you have and add $0$ as a fourth map
Apparently this has a short solution that is non-obvious: let $A$ be a subring of a ring $B$ and let $\beta \in B^\times$. Then every $\alpha \in A[\beta] \cap A[\beta^{-1}]$ is integral over $A$.
@TedShifrin Using induction to prove that when $n$ is an exact power of 2
$$
T(n) =
\begin{cases}
1 & \text{if}\; n=1\,,\\
8T(n/2) + n^2 & \text{if}\; n =2^k, for & k > 1 \,.
\end{cases}
$$
@ÍgjøgnumMeg I think the degree of obviousness (is that even a word?) depends on how you well remember your different characterizations of integral elements
@TedShifrin we're not working w the mercator metric but u get the mercator projecttion by projecting the sphere onto a cylinder tangent along the equator
Let $M$ be a smooth connected manifold and let $p:E \to M$ be a smooth vector bundle with a connection. Is the holonomy group at a point $x$ a closed subgroup of $\mathrm{GL}(p^{-1}(x))$?
it's not the like radial projection but that doesnt matter in this context right the only thing i said is that ur doing transport on the cylinder to get 0 along latitudes
@dondeman Yeah, I didn't check your work, but having all those $k$ in there just makes things harder to sort out.
@Mathein: If you think about the proof of 2-D Gauss Bonnet using differential forms (originally due to Chern, but you can find it in Section 3 of Chapter 3 of my diff geo notes), it's an application of Stokes's Theorem. The general case is the same idea but, obviously, more involved.
@Ted I wasn't even thinking about the proof. I just forgot that the curvature tensor is a 2-form with values in the endomorphisms of the fiber, which is of course just the Lie algebra of GL of the fiber. Since the holonomy group is by definition a subgroup of GL of the fiber, the Lie algebra of the holonomy is not just an abstract Lie algebra but actually automatically embedded in the Lie algebra of GL of the fiber. So it was really a stupid thing
do you know if you have a complex vector bundle over a Riemann surface $E \to \Sigma$, is specifying a map taking sections of E to complex-antilinear E-valued one forms like a d-bar operator enough to induce a holomorphic structure on ur vb
the other way around is true bc if you have a holomorphic vector bundle than you can define $\bar{\partial}$ by just doing it wrt trivializations and noting that transition maps are holomorphic so nothing bad hapens
Right. So, provided you have the integrability condition, can't you turn that around to deduce that the transition functions have to be holomorphic if you're well-defined?
@Jacksoja why do you make that step if you just want to prove something about transitive actions?
@Jacksoja yes!
if you've seen the proof of orbit-stabilizer, try to recall it or look it up or try to redo it yourself. Then think about what happens in terms of group actions
@TedShifrin idk this is like specifying $\nabla$ on a vb so my guess is this isnt true in any nice sense for a general complex vb and a given $\bar{\partial}$-operator
the "proper" way to state the orbit stabilizer theorem is that for a $G$-set $X$ and any $x \in X$, there's a "canonical" isomorphism of $G$-sets (which means a $G$-equivariant bijection) between the orbit of $x$ and $G/\mathrm{Stab}_G(x)$. Can you figure out what that isomorphism is? @Jacksoja
as I said, if you've seen a proof of orbit-stabilizer then you've seen this isomorphism in one form or the other
Question: if $f:A\rightarrow B$ is an injective linear map with $A$ and $B$ vector spaces over some field $K$ and $\tau:B\rightarrow A$ is an onto linear map such that $\tau\circ f=I_A$, then is $\tau$ uniquely determined by $f$?
Consider the function $t(x)$ defined as :
$$ x_1 = x $$
$$x_2 = x $$
$$ x_3 = 2 x^2 $$
$$ x_4 = 4 x^4 + 2 x^2 $$
and for $n > 4 $
$$ x_{n} = \frac { x_{n-1}^2 + x_{n-2}^2 + x_{n-3}^2}{x_{n-2} + x_{n-3} + x_{n-4} } $$
If the sequence converges to a constant then we define $t(x) = \lim x_n $.
...
