Definition of integrable representation of Kac-Moody algebra

I have seen several definitions of integrable representation $ V$ of Kac-Moody algebra $ \mathfrak{g}$ online. Which one is the standard one? Are they actually equivalent?

First one is $ e_i, f_i$ acts nilpotently on $ V$ , where $ e_i, f_i$ are the Chevalley basis of $ \mathfrak{g}$ .

Second one is for any root $ \alpha$ , restriction of $ V$ to the $ sl_2$ corresponding to $ \alpha$ can be integrated to a $ SL_2$ representation.

Third one is $ V$ can be lifted to a representation of the (maximal) simply connected Kac-Moody group whose lie algebra is $ \mathfrak{g}$ .

The case I am interested most is untwisted affine Lie algebra. So feel free to restrict to this case if it helps.

Projection of an invariant almost complex structure to a non integrable one

My apology in advance if my question is obvious or elementary

We identify elements of $ S^3$ with their quaternion representation $ x_1+x_2 i +x_3 j +x_4 k$ . We consider two independent vector fields $ S_1(a)=ja$ and $ S_2(a)=ka$ on $ S^3$ . On the other hand $ P: S^3\to S^2$ is a $ S^1$ -principal bundle with the obvious action of $ S^1$ on $ S^3$ . Then the span of $ S_1, S_2$ is the standard horizontal space associated to the standard connection of the principal bundle $ S^3 \to S^2$ . Then each horizontal space has an almost complex structure $ J$ . This is the standard structure associated to $ S_1, S_2$ coordinate.

Is this structure invariant under the action of $ S^1$ ? If yes, we can define a unique almost complex structure on $ S^2$ which is $ P$ related to the structure on total space. Now is this structure on $ S^2$ integrable?

As a similar question, is there an example of a principal bundle $ P\to X,$ such that $ P$ is a real manifold and $ X$ is a complex manifold and a connection admit an invariant almost complex structure which project to a non integrable structure?

Proving that a function is Riemann-stieltjes integrable

Let $ g$ a increasing function, and $ f$ integrable with respect to $ g$ in $ J=[a,b]$ proof that $ |f|$ is integrable with respect to $ g$

By definition if $ f$ is integrable with respect to $ g$ , for every partition $ P$ of $ J$ and for every $ \epsilon>0$ exists $ I$ in $ R$ such that if $ Q$ refine $ P$ then $ |S(Q,f,g)-I|< \epsilon $ $ |S(Q,f,g)-I|= \sum^{n}_{j=0}f(\lambda_{j})(g(x_{j}-x_{j-1}) $ then

$ \sum^{n}_{j=0}|f(\lambda_{j})|(g(x_{j}-x_{j-1}) \geq \sum^{n}_{j=0}f(\lambda_{j})(g(x_{j}-x_{j-1}) $

but i don’t know how to do

$ \epsilon \geq \sum^{n}_{j=0}|f(\lambda_{j})|(g(x_{j}-x_{j-1}) $

somebody can help me, please?

Prove that $f$ is not Lebesgue integrable

I need a hand with the following exercise:

Prove that $ f: (0,2) \to \mathbb{R}$ given by

$ f(x) = \begin{cases} \frac{1}{x} & 0<x\leq 1 \ \frac{1}{x-2} & 1< x < 2 \end{cases} $

Is not Lebesgue-integrable.

Here are my thought:

We can write $ f$ as $ f = f^+ – f^-$ where

$ f^+(x) = \begin{cases} f(x) & \text{if} \ f(x)>0 \ 0 &\text{otherwise}& \end{cases} $

$ f^-(x) = \begin{cases} -f(x) & \text{if} \ f(x)\le0 \ 0 &\text{otherwise}& \end{cases} $

And by definition $ f$ is integrable if and only if $ f^+$ and $ f^-$ are both integrable, so I just need to prove that $ f^+$ or $ f^-$ is not Lebesgue integrable.

Now, in this particular case $ f^+(x) = \begin{cases} \frac{1}{x} & 0<x\leq1 \ 0 & 1<x<2 \ \end{cases} $ and $ f^-(x) = \begin{cases} 0 & 0<x<1 \ -\frac{1}{x-2} & 1\leq x<2 \ \end{cases} $

So $ f = f^+-f^-$ becomes $ $ f = \frac{1}{x}\chi_{(0,1]}-(-\frac{1}{x-1}\chi_{[1,2)})$ $

thus if I proove that, says, $ \frac{1}{x}$ is not Lebesgue integrable on $ (0,1]$ the problem is solved. But how to prove that?

Show Lebesgue Integrable and Compute the Two Iterated Integrals

(I am working on problems having to do with Fubini’s Theorem)

Given $ α ∈ (0,∞)$ , show that the function $ (x, y) \mapsto e^{−αxy}\cdot sin x$ is Lebesgue integrable on $ (0,∞) × (1,∞)$ . Compute the two iterated integrals and use the result to compute

$ \int_0^{\infty} e^{\alpha x} \frac{sinx}{x}dx $

How do I show the function is Lebesgue integrable? Usually I need to show that the Lebesgue integral is finite… but I am new to having two variables in these problems.

Now, for evaluating the integral. I have evaluated each of them below, then set them equal, as the iterated integrals should be equal. Is that correct?


$ \int_1^{\infty} \int_0^{\infty} e^{-\alpha xy} \cdot sinx dxdy $

$ I = \int_0^{\infty} e^{-\alpha xy} \cdot sinx dx$

Let $ u = e^{-\alpha yx}, du = -\alpha ye^{-\alpha yx}, v = -cosx, dv = sinxdx$ .

$ I = -cosxe^{-\alpha yx}\rvert_0^{\infty} – \alpha y \int_0^{\infty}cosxe^{-\alpha yx}dx $

Let $ u = e^{-\alpha yx}, du = -\alpha ye^{-\alpha yx}, v = sinx, dv = cosxdx$ .

$ I = (0-(-1)(1)) – \alpha y [e^{-\alpha yx}sinx\rvert_0^{\infty} + \alpha y \int_0^{\infty} e^{-\alpha xy} \cdot sinx dx] $

$ I = 1 – \alpha y(0-0) – \alpha^2 y^2 I$

$ I = \frac{1}{1+\alpha^2 y^2}$

Now we have,

$ \int_1^{\infty} \frac{1}{1+\alpha^2 y^2}$

Let $ u = \alpha x, du = \alpha dx$ .

$ = \frac{1}{\alpha} \int_{\alpha}^{\infty} \frac{1}{1+u^2} du = \frac{1}{\alpha} (arctan(\alpha x))\rvert_1^{\infty} = \frac{1}{\alpha} (\frac{\pi}{2} – arctan(\alpha))$


$ \int_0^{\infty} \int_1^{\infty} e^{-\alpha xy} \cdot sinx dydx $

$ =\int_0^{\infty} [\frac{sinx}{-\alpha x} \cdot e^{-\alpha xy}]\rvert_1^{\infty} dx = \int_0^{\infty} \frac{sinx}{-\alpha x} (0 – e^{-\alpha x}) dx = \frac{1}{\alpha} \int_0^{\infty} e^{-\alpha x} \cdot \frac{sinx}{x} dx$

Then I set them equal to evaluate the integral the problem asks for.

$ \frac{1}{\alpha} (\frac{\pi}{2} – arctan(\alpha)) = \frac{1}{\alpha} \int_0^{\infty} e^{-\alpha x} \cdot \frac{sinx}{x} dx$

$ \implies \frac{\pi}{2} – arctan(\alpha) = \int_0^{\infty} e^{-\alpha x} \cdot \frac{sinx}{x} dx$

My issue is that in the problem is is $ \alpha x$ not $ -\alpha x$ .

Prove bounded set implies $L^2$ integrable function?

On a complete and filtered probability space, $ (\Omega, \mathcal{F}, \mathcal{P})$ , I would like to show that a function $ f(t, \omega_1, \omega_2): \mathbb{R} \times \mathbb{R}^2 \to A \in \mathbb{R}$ is $ L^2$ -integrable, i.e. $ \int_0^{\infty}E[f^2(s)]ds < \infty $ if the set $ A$ is bounded. I’m very new to analysis, so I could use some help.