TITLE: Finding the volume of this intersection
QUESTION [1 upvotes]: How to find the volume of this:

The region common to the interiors of the cylinders $x^2+y^2=1$ and
  $x^2+z^2=1$ and the first octant.

I tried finding the volume via double integration and the integration gets too complicated. I want to know how to solve this via triple integration. Usually there are $x,y,z's$ in both equations and that I can solve with triple but I dont know what to do here.
I started of doing $x^2+z^2=x^2+y^2$ so $z=y$ but dont know what to do with this it doesnt look like the line of intersection is $z=y$ in the diagram given.

REPLY [0 votes]: Draw out the surfaces and see that $x$ varies from $0$ to $1$, $y$ varies from $0$ to $\sqrt{1-x^2}$, and $z$ varies from $0$ to $\sqrt{1-x^2}$ as well. There are $5$ other ways to order the integration but aren't as useful as this ordering (check why) and this ordering makes use of squaring the square root
$$V=\int_0^1\int_0^\sqrt{1-x^2}\int_0^\sqrt{1-x^2}dzdydx$$
This is the same as
$$V=\int_0^1\int_0^\sqrt{1-x^2}\sqrt{1-x^2}dydx$$
Which is the area integral of the function $f(x,y)=\sqrt{1-z^2}$, the surface of the "top" cylinder. The other cylinder, $x^2+y^2=1$ serves as the boundary of the integration region. The triple and double integrals are equivalent.
When using the triple integral to find volume, the integrand is simply $1$ and integrate wrt all variables whereas with using the double integral to find volume, the integrand is the function $f(x,y)$ and then integrate wrt $x$ and $y$
As for $z=y$, it appears on the $y-z$ plane as the cylinders make contact at $(0,1,1)$


TITLE: Why the derivative of a function must be equal to zero, on a connected open set, to say that the function is constant?
QUESTION [0 upvotes]: Let $\phi$ a differentiable function on $\Omega \in \mathbb{R}^n$, open connected set.
then if $\frac{\partial\phi}{\partial x_i}=0 \; for \; i=1,..,n$.
we have $\phi = constant$, because $\Omega$ is connected!
I didn't get why $\Omega $ should be connected.

REPLY [2 votes]: If $U$ and $V$ are disjoint open sets, $\phi(x)=1$ for all $x \in U$ and $\phi(x)=0$ for all $x \in V$ then all the partial derivatives are $0$ but $\phi$ is not a constant function on $U \cup V$.

REPLY [1 votes]: Since $\Omega$ is open and connected, any two points $x,y \in \Omega$ can be
joined by a path with a finite number of segments that are parallel to the axes.
Then the mean value theorem shows that $\phi(x)=\phi(y)$ and hence $\phi$ is constant
on $\Omega$.


TITLE: When is the Ad (Adjoint Representation) Morphism a Closed Map
QUESTION [7 upvotes]: Given a Lie group $\mathfrak{G}$ with finite centre and with Lie algebra $\mathfrak{g}$, I am looking at a simple proof that negative definite Killing form implies compactness. This proof is given here. A sketch of the proof is as follows:

The author claims that $\mathrm{Ad}:\mathfrak{G}\to GL(\mathrm{Lie}(\mathfrak{G})$ is a closed map;
The Killing form $K:\mathfrak{g}\times\mathfrak{g}\to\mathbb{R}$ is $\mathrm{Ad}$-invariant, so a negative definite $K$ allows us to define an inner product $-K$such that $\mathrm{Ad}(\gamma)$ is orthogonal wrt $-K$ i.e. $\mathrm{Ad}(\mathfrak{G})\subseteq SO(\dim \mathfrak{g}, -K)$.
$SO(\dim \mathfrak{g}, -K)$ being compact, and $\mathrm{Ad}(\mathfrak{G})$ a closed subgroup by 1., we conclude that $\mathrm{Ad}(\mathfrak{G})$ is itself compact;
$\mathfrak{G}$, being an $M$-fold cover of $\mathrm{Ad}(\mathfrak{G})$ (where $M$ is finite by dint of the finite centre), is thus also compact.

Crucial to this proof is the assertion that  $\mathrm{Ad}(\mathfrak{G})$ is closed in $SO(\dim \mathfrak{g}, -K)$. I can prove this given nondegeneracy of the Killing form, (e.g. with Lemma 1 of G. Hochschild, "Complexification of Real Analytic Groups") but the author of the first document I linked seems to be saying that this is a much more general and well known property of $\mathrm{Ad}$. What am I missing here?: I think I'm making this harder than it should be through overlooking a simple fact.  So, as in my title:
When is $\mathrm{Ad}:\mathfrak{G}\to GL(\mathrm{Lie}(\mathfrak{G}))$ a closed map and why?
Edit: It seems that this is not as trivial as I thought. Hence answers less than a full answer are helpful and acceptable to me. For example, interesting counterexamples (showing when $\mathrm{Ad}:\mathfrak{G}\to GL(\mathrm{Lie}(\mathfrak{G})$ is not closed) or sufficient conditions for it to be closed (such as semisimplicity of $\mathfrak{G}$).

REPLY [5 votes]: Let $G$ be a connected Lie group. Equivalences:

(i) every linear representation of $G$ has a closed image
(ii) $\mathrm{Hom}(G,\mathbf{R})=0$
(iii) $G/\overline{[G,G]}$ is compact.

Here homomorphisms are meant continuous, and reps are in $\mathbf{GL}_n(\mathbf{R})$ for $n$ not fixed, or equivalently in $\mathrm{GL}_n(\mathbf{C})$ in view of the closed inclusions $\mathrm{GL}_n(\mathbf{R})\subset \mathrm{GL}_n(\mathbf{C})\subset \mathrm{GL}_{2n}(\mathbf{R})$. Note that the equivalence between (ii) and (iii) is obvious.
That (i) implies (ii) is clear since if $\mathrm{Hom}(G,\mathbf{R})\neq 0$, $G$ admits $\mathbf{R}$ as a quotient, and we can use a 2-dimensional complex representation of $\mathbf{R}$ whose image is dense in a 2-dimensional real torus.
The implication (ii)$\Rightarrow$(i) goes by reduction to the semisimple case.
Thus first assume that $G$ is semisimple. Fix a representation $\rho:G\to\mathrm{GL}_n(\mathbf{C})$. In the latter, the image of the representation is a (a priori not closed) Lie subgroup $H_1$, with Lie algebra $\mathfrak{h}$ (a real Lie aubalgebra of $\mathfrak{gl}_n(\mathbf{C})$.
Let $H_2$ be the closure of $H_1$ (in the real Zariski topology, or in the real topology, it does not matter). Then $H_2$ normalizes $\mathfrak{h}$ and $H_1$. Since up to finite index, the automorphisms of $H_1$ are inner, and since $H_2$ is connected, we deduce that $H_2=H_1C$ where $C$ is the centralizer of $H_1$ in $H_2$. Decompose $\mathbf{C}^n=\bigoplus V_i$, a sub of $H_1$-irreducible complex subspaces. Then since $C$ is in the closure of $H_1$, it preserve each $V_i$. Since the $H_1$-representation on the complex space $V_i$ is irreducible, its commutant is reduced to scalars. Moreover, since the restriction of $H_1$ on $V_i$ has determinant 1, it is also the case for $C$. We deduce that $C$ is a finite abelian group. Thus $H_1$ has finite index in $H_2$. Since $H_1$ is $\sigma$-compact, a Baire argument then shows that $H_1$ has non-empty interior in $H_2$, and hence $H_1$ is the unit component of $H_2$ in the real topology. Thus $H_1$ is closed.
(Another argument goes by using the fact that perfect Lie subalgebras of $\mathfrak{gl}_n(\mathbf{R})$ are always Lie algebras of some Zariski closed subgroup of $\mathrm{GL}_n(\mathbf{R})$.)
Now assume $G$ arbitrary satisfying (ii). Write $G=RS$ with $R$ the connected solvable radical and $S$ a semisimple Levi factor. A simple argument
 based on taking the complexification at the Lie algebra level shows that $[G,R]$ is normal and that each representation of $G$ maps $[G,R]$ to unipotents. Thus let $\rho$ be a representation of $G$, say in $\mathrm{GL}_n(\mathbf{R})$, and let $H$ be its image. Then $N=\rho\Big(\overline{[G,R]}\Big)$ is a connected, unipotent normal subgroup of $H$. Being a unipotent connected Lie subgroup, it is necessarily closed in $\mathrm{GL}_n(\mathbf{R})$, actually Zariski closed. If $M$ is the normalizer of $N$, then $H\subset M\subset  \mathrm{GL}_n(\mathbf{R})$ and $M$ is Zariski closed. Moreover $M/N$ stands a Zariski closed subgroup of $\mathrm{GL}_m(\mathbf{R})$ for some $m$ (indeed, if $M=\mathbb{M}_\mathbf{R}$ and $N=\mathbb{N}_\mathbf{R}$ for some $\mathbf{R}$-subgroups of $\mathrm{GL}_n$, then $\mathbb{M}/\mathbb{N}$ is a linear algebraic group and hence is isomorphic to an $\mathbf{R}$-closed subgroup in some $\mathrm{GL}_m$, and $\mathbb{M}_\mathbf{R}/\mathbb{N}_\mathbf{R}\to (\mathbb{M}/\mathbb{N})_\mathbf{R}$ is a closed map in the real topology.)
Therefore, we are reduced to the case when $[G,R]=1$. In this case $G=RS$, with $S$ semisimple Levi factor, $R$ abelian, and $[R,S]=1$. Since every linear representation of $S$ factors through some finite index subgroup of its center, it is no restriction to assume that $S$ has a finite center, and in particular $S$ is closed. By (ii), we thus deduce that $R$ is compact. Hence since by the first case, $\rho(S)$ is closed, we deduce that $\rho(G)$ is closed as well.


TITLE: Truth table for $(p\land q)\rightarrow r$
QUESTION [0 upvotes]: I want to know if I'm on the right track with my truth table for
(p AND q) implies r....
p | q | r | p AND q | (p AND q) --> r
T   T   T      T        T
T   T   F      T        F
T   F   T      F        T
T   F   F      F        F
F   T   T      F        T
F   T   F      F        F
F   F   T      F        T
F   F   F      F        F

I don't really understand the implication rule when used like this so if anyone can give me a further breakdown that would be great
Thanks!

REPLY [1 votes]: Let us consider the $r$ and the $(p\land q)$
Just remember the rules for implication
If we have $P\to Q$ (these are just dummy variables to show you the relationship), the possible values are as follows:

The way you wrote your table is correct.
In the first row we have r=T, p AND q = T, and as we know $T\to T = T$, hence you get T.
If we consider the second row, we have r=F, p AND q = T. But remember we are doing $(p\land Q)\to r$, so we have $T\to F$, which is $F$ by the table, and thats how you get F in row 2. Just repeat this for all the rows.

REPLY [0 votes]: $P \rightarrow Q$ is always True, except when $P$ is True and $Q$ is False.
So, in your case, whenever $p \land q$ is false (rows 3 through 8), $(p \land q) \rightarrow r$ is True


TITLE: Is this an inconsistence in the usage of the space of all continuously differentiable functions $C^{n}(X, E)$?
QUESTION [0 upvotes]: My textbook defines the space of all continuously differentiable functions $C^{n}(X, E)$ as follows:

As such, $f \in C^{n}(X, E)$ IFF the $n$-th derivative of $f$ is continuous on $X$. I found that this definition coincides with one given by Wikipedia's page.
Roll's theorem is stated in my textbook as:

From the definition, I got that $f \in C([a, b], \mathbb{R})$ IFF $f'$ is continuous on $[a,b]$, which is stronger than "$f$ is differentiable
on $(a,b)$". This causes me confusion.
The authors already have $f \in C([a, b], \mathbb{R})$, but they add that $f$ is differentiable on $(a,b)$. Is it a redundancy?
Please elaborate on this notion!

REPLY [1 votes]: You're confusing $C^1$ with $C$. $C$ just means $f$ is continuous and places no requirements on differentiability. $C^1$ means differentiable with continuous derivative.

REPLY [1 votes]: $C$ is $C^{0}$ and the zeroth derivative of a function is the function itself.


TITLE: Maximum of $\vert f(z)\vert$ on $|z|\leqslant1$, where $f(z)= 1+z^2$
QUESTION [1 upvotes]: Let $f(z)=z^2+1$. Determine the maximum of $\vert f(z)\vert$ on $\overline{D_1(0)}$

First I want to use the maximum theorem for a complex holomorphic function. But I thought this is an easier way:
$$\vert f(z)\vert = \vert 1+z^2\vert \leq 1+ \vert z^2\vert = 1+ \vert z\vert^2$$
Due to the fact that this function is defined on $\overline{D_1(0)}$ one can say, that $\vert z\vert \leq 1 \forall z\in\mathbb{C}$ $$\Rightarrow \max\limits_{\overline{D_1(0)}} \vert f(z)\vert = 2$$
Is that correct? Thank you!

REPLY [2 votes]: After you show that $$|f(z)| \leq 2$$
Show a point $z$ that makes $|f(z)|$ attain that value, in particular we can let $z=1$.
Hence the maximum value is indeed $2$.


TITLE: Is the Product of Banach Spaces a Banach Space?
QUESTION [0 upvotes]: Let $X$ and $Y$ be two Banach spaces (not necessarily possessing the same norm). The product space $X×Y=Z$ is given the max norm, i.e. $\max(\Vert x\Vert, \Vert y\Vert)$, where $x$ is given the norm of $X$ and $y$ is given the norm of $Y$, and where $(x,y)$ is an element of $Z$. Prove that the product space $Z$ is Banach.

REPLY [1 votes]: $z_n=(x_n,y_n)$ is a cauchy (convergent) sequence in $Z$ iff $(x_n)$ and $(y_n)$ both are cauchy (convergent).