Josué

I want an algebra with an inner product, in other words.

I guess I mean multiplication in the algebra sense.

etc.

I know that a normed linear space with a "multiplication" operation is called a normed algebra.

I know that a linear space with a "multiplication" operation is called an algebra.

Let me rephrase: is there a name for an inner product space with a "multiplication" operation?

Is there a name for a field with an inner product?

another chess match anyone? lichess.org/gKJURUfo

chess anyone? lichess.org/xG5EZakd

@LeakyNun I love pure maths, despite everything I do being applied.

@CaptainAmerica16 I usually tell the curious that I wanted a better understanding.

I ended up doing signal processing.

@CaptainAmerica16 That's exactly what I did.

@LeakyNun Does this affect proofs in topology that use the axiom of choice, like Tychonoff's theorem?

What does it mean that the axiom of choice fails in the category of, e.g., topological groups?

hi

@TedShifrin I looked it up, and it taught me something new, but the notation of a Hölder space is $C^{k,\alpha}$, where $k\geq0$ is an integer and $0<\alpha\leq1$.

Hello! $C^n$, where $n\in\mathbb N$, is the set of continuous, $n$-times differentiable functions. However, while reading a publication, I saw $C^\beta$, where $\beta\in\left[1,2\right)$. What in the world does this mean?

Is $(C([-1,1]),\|\cdot\|_{L^2})$ complete?

I think I got it using the Fourier inversion formula.

Could someone please hint how I may show $f,\hat f\in L^1(\mathbb R)$ implies $f\in L^p(\mathbb R)$ for all $p\in[1,\infty]$?

I suppose that this is true if $X$ and $Y$ are isomorphic in some sense.

Let $X$ be a Banach space, and let $A\subset L(X,X)$ be the set of invertible bounded linear operators on $X$. I know that $A$ is open in $L(X,X)$ with respect to the operator norm and can prove it. However, If $X$ and $Y$ are Banach spaces, does it still hold that $A\subset L(X,Y)$ is open?

@TedShifrin LOL. I think I need a vacation, hahahaha. Indeed, it is complete by definition.

@TedShifrin I tried, but I am hopelessly stuck. :(

Is there a shorter way to prove Hilbert spaces are complete?

Every Hilbert space has an orthonormal basis. Therefore, every Hilbert space has a unitary on it taking values on some $\ell^2$ space. Since $\ell^2$ is complete, every Hilbert space is complete.

@TedShifrin I used the quotient rule, but I got a ridiculously long expression. -.-

May I say that $\displaystyle\frac{\sqrt{-1}}2\partial\bar\partial\log(1+z_1\bar z_1+z_2\bar z_2)=\frac{\sqrt{-1}}2\partial\frac{z_1d\bar z_1+z_2d\bar z_2}{1+z_1\bar z_1+z_2\bar z_2}?$

Hey, @TedShifrin! I'm now looking at $\mathbb P^2$ with the Fubini-Study metric $g$. I know that the associated $(1,1)$-form of $g$ is $\omega=\sqrt{-1}/2\partial\bar\partial\log(1+z_1\bar z_1+z_2\bar z_2)$ for affine coordinates $z_1,z_2$. Any hints on how to make sense of $\omega^2$?

@TedShifrin I got it, Ted. Thanks!

@TedShifrin I will try that. Thank you so much. :)

@TedShifrin Could you please elaborate a bit more on why only the $dz_i\wedge d\bar z_i$ terms are picked? I intuitively agree with this statement but I am unable to pinpoint why. For example, up to constants, $\text{Ric}\wedge\omega^{n-1}=(\sum r_{i\bar j}dz_i\wedge d\bar z_j)\wedge(\sum dz_i\wedge d\bar z_i)^{n-1}$.

@TedShifrin Thank you, Ted. I am actually carrying out the computations myself. So, for example, if $\omega=\sqrt{-1}/2\sum_{i=1}^ndz_i\wedge d\bar z_i$ and $\text{Ric}=\sqrt{-1}\sum_{i=1}^nr_{i\bar i}dz_i\wedge d\bar z_i$, then I would like to actually compute $\text{Ric}\wedge\omega^{n-1}$.

How does that follow?

someone please, please help me keep my sanity

$\int_{\Delta^2}1/(|z|^{2a}+|w|^{2b})$. For which $a,b>0$ is this integrable? How may I determine this?

I know all norms on finite-dimensional vector spaces are equivalent. So I guess I can say $|(z_1,z_2)|=|z_1|+|z_2|$ on $\mathbb C^2$.

How is the absolute value of a point in $\mathbb C^2$ defined?

Is $\Delta^n$ the polydisc?

Is anyone here good with algebraic geometry?

Good evening, gentlemen. I was told that $\text{dim }N^1(\mathbb P\times\mathbb P)_{\mathbb R}=2$. How may I show that?

Who own's Huybrechts' Complex Geometry?

how to compute $$\int_{\mathbb P^1}c_1(L)$$?

How do I compute $\int_{\mathbb{P}^1}c_1(L)$?

I observed that $\tilde{\mathbb C}^2=\{((z_1,z_2),[z_1,z_2])\}$ and $E=\{(0,0)\}\times\mathbb P^1$.

Can I now extend $f$ across $E$?

So lift $f$ to its "blow up", i.e., $f:\tilde{\mathbb C}^2\setminus E\to\mathbb P^1$, where $\tilde{\mathbb C}^2=\{((z_1,z_2),[l_1,l_1])\in\mathbb C^2\times\mathbb P^1:z_1l_2=z_2l_1\}$ and $E=\pi^{-1}(\{(0,0)\})$, where $\pi:\tilde{\mathbb C}^2\to\mathbb C^2$ is the first projection map.

hello