arctic tern

if you want negative points then 0.999=1

"does it"? the fact you're talking about is called taylor's theorem

err, right

you mean does Taylor's Theorem have a proof? sure, google away.

f(x+h)=f(x)+f'(x)h+f''(x)h^2 would be a 2nd order approximation

with division, if you want to approximate 1776 with a multiple of 10, you have 1770, and the remainder (difference between estimate and true value) is 6

@Kirill remainder of taking away the approximation

@Kirill r stands for "remainder"

if r=f(x+h)-f(x)-f'(x)h then r(h)~f''(x)h^2 no? (also, shouldn't r depend on x too?)

ah, I see, you're viewing hexagons as 2D, not 1D

how do points on the line segments have nbds homeomorphic to the closed half-disk?

@AkivaWeinberger homeomorphic or homotopy equivalent?

ah, I didn't see the word hyperplane

if so then Z must have codimension 1 in order for it to be the kernel

extend basis from Z to X, use to define f

for instance when n=1 we can write H=C+jC, with C acting on the right. then the original left multiplication by i will act as the nonscalar matrix diag(i.-i).

but Sp(n)Sp(1) intersect U(2n) (where U(2n) acts on H^n=C^2n from the left) will just be Sp(n) rather than Sp(n)U(1) since right multiplication of C^2n by i will not match the original left multiplication by i

Since Sp(n) and Sp(1) commute in Sp(n)Sp(1), to find the centralizer of i (the element of Sp(1)) it suffices to compute its centralizer purely in Sp(1), which will be U(1)

yes

if your complex structure is right multiplication of H^n by i, then that's an element of Sp(n)Sp(1), and its centralizer is Sp(n)U(1)

multiplication of H on the right by i doesn't match multiplication on the left by i

@Daminark aight

not sure what almost complex structure you're talking about, unless you mean multiplication of C^2 by i as usual

Sp(n) is a subset of U(2n) where both are acting on H^n from the left (to commute with scalars coming from the right), no?

multiplying elements of H^n on the left by exp(i phi) is different from multiplying by it on the right, unless exp(i phi) is real

@Kirill Akiva's integrand is exp(exp(i*theta)), which is different from what you're talking about

because $(a+b)c=ac+bc$ and $a^ra^s=a^{r+s}$

for example, $(e^{2ix}+3e^{-ix})e^{-5ix}=e^{(2-5)ix}+3e^{(-1-5)ix}$

@Kirill $\int \sum=\sum\int$ and $\int_{-\pi}^{\pi} e^{i kx}dx=2\pi \delta_k$

@Danu I guess it's standard for Sp(n)Sp(1) in GL(4,R) to act via Sp(n) on the left of H^n (=R^4n) and via Sp(1) on the right, so it's Sp(n)xSp(1) mod +/-(1,1). On the interpretation that U(2n) acts from the left on C^2n=R^4n, that should make Sp(n) a subgroup of U(2n) and Sp(1) intersect it in +/-1, in which case I'd expect the intersection of Sp(n)Sp(1) and U(2n) to just be Sp(n)...

@Danu Doesn't Sp(n) contain Sp(1)?

riemann surfaces & algebraic geometry over C?

SL(n,Z) is normal in SL(n,Q) because it is normal in the supergroup diag(Q)*SL(n,Z), no?

why say "generated by" when you gave a whole coset? :P

a finite set of coset representatives would have finitely many primes in their denominators, and any SL(n,Q) matrix involving other primes in the denominator would not be in the group generated by the coset reps and SL(n,Z)

@TobiasKildetoft not sure what you mean

@Kirill I can't understand what you're saying.

@Kirill don't you already know $\langle au,v\rangle=\overline{a}\langle u,v\rangle$ and $\langle u,av\rangle=a\langle u,v\rangle$? (well, that's the physicist convention, the mathematician's convention is often the reverse)

@LucasHenrique big

$a\cdot\langle u,v\rangle=\langle u,v\rangle$ is not true unless $a=1$ or $\langle u,v\rangle$. and in any case $a=\overline{a}$ if and only if $a\in\Bbb R$

@Kirill If you had it in both arguments it'd be more or less the same as in no arguments and just took the conjugate of the output, since $\overline{z}\,\overline{w}=\overline{zw}$

for example, f(z,w)=zw is bilinear on C^1, whereas $\overline{z}w$ is sesquilinear

So, they are indeed different.

With a bilinear form, <a,a> may not be real. On the other hand, with a sesquilinear form, <a,a> is always real. You can use the standard inner product to measure size, after all.

@Kirill I don't understand what you're asking.

just set $u=\varphi^{\ast}(v)$ then

@Kirill do you agree $\langle \varphi^{\ast}(v),u\rangle=\langle v,\varphi(u)\rangle$ for any $u,v,\varphi$ by definition of adjoint?

indeed, $\langle \varphi^{\ast}(v),u\rangle=\langle v,\varphi(u)\rangle$ for any $u,v,\varphi$, by definition of adjoint

@Kirill $\|\varphi^{\ast}(v)\|^2=\langle \varphi^{\ast}(v),\varphi^{\ast}(v)\rangle=\langle v,\varphi(\varphi^{\ast}(v))\rangle$