TITLE: An exercise in Artin's Algebra
QUESTION [2 upvotes]: The exercise is from Chapter 9 "Linear group".

Q4.8b: Let $W$ be the real vector space of Hermitian 2 x 2 matrices. Prove that the function $\langle A,A'\rangle =\det(A+A')-\det(A)-\det(A')$ is a bilinear form on $W$, and that its signature is $(3,1)$.

Here signature $(3,1)$ means the matrix of that bilinear form is $$\begin{pmatrix}
  1 & 0 & 0 & 0 \\
  0 & 1 & 0 & 0 \\
  0 & 0 & 1& 0  \\
 0 & 0& 0 & -1
 \end{pmatrix}\;.$$
(Indeed, I have calculated the signature should be $(1,3)$, i.e. the matrix of the form should be $$\begin{pmatrix}
  1 & 0 & 0 & 0 \\
  0 & -1 & 0 & 0 \\
  0 & 0 & -1& 0  \\
 0 & 0& 0 & -1
 \end{pmatrix}$$ so I'm thinking about the question missed a negetive sign in the bilinear form i.e. it should be $\langle A,A'\rangle =-[\det(A+A')-\det(A)-\det(A')]$, then we can have signature $(3,1)$, am I right or not?)

REPLY [1 votes]: Easiest to calculate the quadratic form associated. This is just
$$\frac{1}{2}B(A,A)=\det A$$
For an arbitrary element of your vector space write it in the form
$$A=\begin{pmatrix}x+y&z+iw\\
z-iw&x-y\end{pmatrix}$$
Then $$\det A=x^2-y^2-z^2-w^2$$


TITLE: Compatibility of $f^*$ and $\mathbb{V}$.
QUESTION [0 upvotes]: Let $Y$ be a scheme. For a locally free $\mathcal{O}_Y$-module $\mathcal{E}$ of rank $n$, the scheme $\mathbb{V}(\mathcal{E})$ is defined as $\mathbb{V}(\mathcal{E}) := \mathbf{Spec}(S^\bullet\mathcal{E})$.
Now if $f: X \rightarrow Y$ is a morphism of schemes, we can consider the locally free $\mathcal{O}_X$-module $f^*\mathcal{E}$, and the associated scheme $\mathbb{V}(f^*\mathcal{E})$.
Are $f^*$ and $\mathbb{V}$ compatible in the sense that there exists a (unique?) morphism $\mathbb{V}(f): \mathbb{V}(f^*\mathcal{E}) \rightarrow \mathbb{V}(\mathcal{E})$ that commutes with the canonical maps $p_X:\mathbb{V}(f^*\mathcal{E}) \rightarrow X$ and $p_Y:\mathbb{V}(\mathcal{E}) \rightarrow Y$? I.e. does
$$p_Y\circ\mathbb{V}(f) = f \circ p_X$$
hold?
My problem is that I do not know how to relate the sheaves $f^*\mathcal{E}$ and $\mathcal{E}$, as both live on different spaces, let alone the schemes defined by those.

REPLY [2 votes]: Such a morphism is not unique. For instance, if $X = Y = Spec(k)$ and $f$ is the identity, then $V(E) = V(f^*E)$ is just an affine space, and any its automorphism satisfies the required compatibility.
On the other hand, there is a canonical choice of a morphism induced by the natural isomorphism of sheaves of algebras
$$
f^*(S^\bullet E) \cong S^\bullet(f^* E).
$$


TITLE: Confused solving quasilinear PDE
QUESTION [2 upvotes]: The equation to solve is:
$\displaystyle x\frac{\partial u}{\partial x}+ \displaystyle y\frac{\partial u}{\partial y}+ \displaystyle z\frac{\partial u}{\partial z}= \displaystyle xyz$
Applying the Lagrange-Charpit equations to $x$ and $y$, and $y$ and $z$ we trivially get $\frac{x}{y}=C_1$ and $\frac{z}{y}=C_2$. However I have a problem solving
$(1) \displaystyle \frac{x}{dx}=\frac{xyz}{du}$
This leads to
$\displaystyle xdu=xyzdx$, and
$\displaystyle du=yzdx$, integrating
$\displaystyle u+C_3=xyz$
This would mean that the solution to the original problem is
$F(\displaystyle \frac{x}{y},\frac{z}{y}, \displaystyle u-xyz)=0$ with an arbitrary $F$.
And indeed $\displaystyle u=xyz$ satisfies $(1)$. But the third function in the actual solution should be $3u=xyz$ which in turn does not satisfy $(1)$.
What am I missing?

REPLY [0 votes]: Your Lagrange equations are statements about the relation of x,y,z,u along a curve in 4-space. This is somewhat hidden in the notation, but is brought out clearly in the solution by doraemonpaul. So when you say that $x/y=C_1$, that means along one of these curves; for the pde, x and y are still going to be independent variables. So when you say that your equation (1) leads to $du = yzdx$, that is still a statement along the same curve. So your deduction $u+C_3 = xyz$ is not correct. You could write instead $du = yzdx=\frac{x}{C_1}\frac{C_2x}{C_1}dx$, and integrate to get
$u = \frac{1}{3}\frac{x}{C_1}\frac{C_2x}{C_1}x+C_3 = \frac{1}{3}xyz+C_3$.
In general I think it is better to include the $t$ parameter as doraemonpaul did, because the idea of the method of characteristics is to use these parametrized curves to transfer Cauchy data into the space.


TITLE: What are some $e^x$ identities to help me complete this proof?
QUESTION [0 upvotes]: While working on a homework problem, I am hitting a point where I can kind of see where I am headed, but I'm not sure how to resolve a couple of items.
I am currently sitting at the following expression:
$$\frac{1}{2\sqrt{2\pi y}} e^{\frac{-\mu^2}{2}}e^{\frac{-y^2}{2}}(e^{\mu\sqrt{y}} + e^{-\mu\sqrt{y}})$$
and need to get to:
$$\sum_{k = 0}^\infty \frac{(e^{\frac{-\mu^2}{2}})(\frac{\mu^2}{2})^k}{k!} * \frac{1}{2^{\frac{2k + 1}{2}}\gamma(\frac{2k + 1}{2})}y^{\frac{2k + 1}{2}-1}e^{\frac{-y}{2}}$$
The second part is a Gamma distribution $(\frac{2k+1}{2}, 2)$
I can see that the first two listed exponentials are likely to move to the gamma distribution, but is there some identify that will allow me to take $(e^{\mu\sqrt{y}} + e^{-\mu\sqrt{y}})$ and change it in order to cut out the even $k$ values of the infinite sum?

REPLY [2 votes]: For the
$(e^{\mu\sqrt{y}} + e^{-\mu\sqrt{y}})$
question,
note that this is essentially the
$\cosh$ function defined by
$\cosh(x)
=\frac12(e^x + e^{-x})
$.
This cancels out the odd terms,
because
$\dfrac{x^{2k+1}}{(2k+1)!}
+\dfrac{(-x)^{2k+1}}{(2k+1)!}
=0
$.
Similarly,
to cancel out the
even terms, use the
$\sinh$ function defined by
$\sinh(x)
=\frac12(e^x - e^{-x})
$.
This cancels out the even terms,
because
$\dfrac{x^{2k}}{(2k)!}
-\dfrac{(-x)^{2k}}{(2k)!}
=0
$.


TITLE: Why this parametric function for an explicit cubic form?
QUESTION [0 upvotes]: I'm trying to understand a statement here. It states that the parametrisation of a cubic function is:
$$\begin{align}
x(t)&=a_xt^3+b_xt^2+c_xt+d_x\\
y(t)&=a_yt^3+b_yt^2+c_yt+d_y\\
\end{align}$$
Or in compact from
$$\begin{align}
P(x,y)&=at^3+bt^2+ct+d\\
\end{align}$$
being a, b, c and d 2d vectors.
But I can't get it. How do they get to this conclusion?

REPLY [0 votes]: Maybe that you are misinterpreting what the link say.  The text is:

A parametric cubic curve in 3D is defined by: .....

so these equations are the definition of a parametric cubic curve (in $\mathbb{R}^3$), not the parametric representation of a cubic as:
$$y=ax^3+bx^2+cx+d \qquad (1)$$, that is a curve in $\mathbb{R}^2$.

Added after the change in the question.
In $\mathbb{R}^2$ the cubic $(1)$ can be expressed in the parametric form:
$$\begin{align}
x(t)&=a_xt^3+b_xt^2+c_xt+d_x\\
y(t)&=a_yt^3+b_yt^2+c_yt+d_y\\
\end{align}\qquad (2)$$
simply using $t=x$ and
$a_x=b_x=d_x=0 \quad,\quad c_x=1\quad,\quad a_y=a \quad,\quad b_y=b\quad,\quad c_y=c\quad,\quad d_y=d$
but note that, in general, a cubic expressed as $(2)$ with the coefficients of $t^2$ and $t^3$  not null, cannot be write in the form $(1)$.


TITLE: number of unordered pairs of integers from 1 to N which are coprime
QUESTION [0 upvotes]: How many unordered pairs of integers from 1 to N are coprime?
FOR N=4,ans is 6
.for n=3,ans is 4.
FOR N=4 I MADE THESE PAIRS (1,4) (3,4) (2,1)(2,3) (1,3) SO WHAT IS THAT I AM MISSING ?
FOR N=3 I MADE THESE PAIRS (1,3) (2,3) (1,2) BUT ANSWER IS 4.
how will the answer differ if we say ordered pair of integers?

REPLY [3 votes]: As @lulu mentions in the comments, $1$ is coprime with $1$, so that $\{ 1,1 \}$ is such a pair.
More generally, for each $n \le N$ there are $\varphi(n)$ unordered pairs of the form $(k,n)$ with $k \le n$ and $k,n$ coprime, where $\varphi$ denotes Euler's totient function. So in general the number of unordered coprime pairs of numbers less than or equal to $N$ is
$$\sum_{n=1}^N \varphi(n)$$
This agrees with your answers:

$N=3$: $\varphi(1)+\varphi(2)+\varphi(3) = 1+1+2=4$.
$N=4$: $\varphi(1)+\varphi(2)+\varphi(3)+\varphi(4) = 1+1+2+2=6$.


TITLE: Does $\bigcup\emptyset$ equal $\bigcup\{\emptyset\}$?
QUESTION [6 upvotes]: For any collection of sets $A = \{A_i\ : i \in I \}$, define
$$\bigcup A = \bigcup_{i \in I} A_i$$
Question: Is the following true?
$$\bigcup \emptyset = \bigcup \{\emptyset \}$$
The right-hand side, using the definition, is simply the empty set.  However, I'm not sure how I'm supposed to interpret the left - hand side.  I'm guessing the fact that the left-hand side is the empty set follows "vacuously", but I'm not too certain.
Can somebody help me understand?

REPLY [1 votes]: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
$To expand on the other answers, here is a slightly more detailed proof that both sets are empty, but for different reasons.
For the first set, we calculate its members $\;x\;$ as follows:
$$\calc
x \in \bigcup \emptyset
\calcop{\equiv}{definition of $\;\bigcup\;$}
\langle \exists V : V \in \emptyset : x \in V \rangle
\calcop{\equiv}{definition of $\;\emptyset\;$}
\langle \exists V : \text{false} : x \in V \rangle
\calcop{\equiv}{logic: simplify}
\text{false}
\endcalc$$
For the second set, we similarly calculate for all $\;x\;$:
$$\calc
x \in \bigcup \{\emptyset\}
\calcop{\equiv}{definition of $\;\bigcup\;$}
\langle \exists V : V \in \{\emptyset\} : x \in V \rangle
\calcop{\equiv}{definition of $\;\{\ldots\}\;$}
\langle \exists V : V = \emptyset : x \in V \rangle
\calcop{\equiv}{logic: one-point rule}
x \in \emptyset
\calcop{\equiv}{definition of $\;\emptyset\;$}
\text{false}
\endcalc$$
Therefore both sets are empty, and therefore they are equal.


TITLE: Is the pullback of differential geometry also a pull back in the sense of category theory?
QUESTION [7 upvotes]: I do think that the pullback of differential geometry is also a pullback in the sense of category theory, however I did have some trouble completely justifying this thought. This is were I got:
Lets consider a vector field $\eta : Y \rightarrow T(Y)$ and a diffeomorphism $f: X \rightarrow Y$. Then from differential geometry the pullback can be written as,
$$ f^*(\eta)(x) = Tf^{-1}\circ \eta \circ f(x).$$
Thus the commutative diagram becomes:

Now since $Tf^{-1}$ is invertible, we can use $Tf$ to write commutative diagram as a fiber product. From category theory we can complete the diagram of the vector field $\eta: Y \rightarrow T(Y)$ and the tangent map $Tf: T(X)\rightarrow T(Y)$ to obtain

Now due to the uniqueness of the fiber product up to isomorphism we see that $Y \times_{T(Y)} T(X)$ equals $X$ . Also $\eta^*(Tf)$ should equal $f$ (up to isomorphism). But then $f^*\eta$ should equal $Tf^*\eta$, however the notation $f^*\eta$ does not make much sense to me, is this just a short hand? If this isn't the case can someone give me somewhat more information about $ f^*\eta$ in perspective of category theory?
Thanks in advance.

REPLY [7 votes]: The right way to think about this is that the answer to your question is no.  It may be possible to write the pullback of a vector field along a diffeomorphism as a pullback in the sense of category theory, but this is basically a coincidence and is not normally a helpful way to understand the concepts.
The word "pullback" is used very broadly to describe, well, pulling things back.  In other words, whenever you have a map $f:X\to Y$, and there is a natural way to turn some structure on $Y$ into a structure on $X$, you can describe that as a "pullback".  Pullbacks in the sense of category theory are one example of such a thing (where the structure on $Y$ is a map from some object to $Y$), and pullbacks of vector fields are another (where the structure on $Y$ is a vector field; notably, though, you can only perform such a pullback if $f$ is an open immersion).


TITLE: Show that the following is true for a bounded sequence in $W^{1,p}(I)$
QUESTION [3 upvotes]: Let $I=(0,1)$. Assume that $u_n$ is a bounded sequence in $W^{1,p}(I)$ with $1 \lt p \le \infty$.

Show that there exist a subsequence $(u_{n_k})$ and some $u$ in $W^{1,p}(I)$ such that $||u_{n_k}-u||_{L^{\infty}} \to 0$.
Show that $u_{n_k}'\rightharpoonup u'$ weakly in $L^{p}(I)$ if $1 \lt p \lt \infty$
Show that $u_{n_k}'\rightharpoonup^{*} u'$ in $\sigma(L^{\infty},L^1)$ if $p=\infty$

For (1) $W^{1,p}(I)$ is compactly embedded in $C(\bar{I})$ for $1 \lt p \lt \infty$. Thus the identity map being compact has a convergent subsequence in $C(\bar{I})$. I am not able to show that $u \in W^{1,p}(I)
 $
For (2),do I have to use that $C_c^{\infty}(I)$ is dense in $L^{P}(I)$??
For (3), How do I show this??
Thanks for the help!!

REPLY [3 votes]: Here is another solution. Let $1 < p \le \infty$.
Since $(u_n)_{n \ge 1}$ in bounded in $W^{1,p}(I)$, which is compactly embedded in $C(\overline I)$, there exist a subsequence $(u_{n_k})_{k\ge 1}$ and $u \in C(\overline I)$ such that $\| u_{n_k} - u \|_{L^\infty} \to 0$.
The sequence $(u'_{n_k})_{k \ge 1}$ is bounded in $L^{p}(I)$, hence there exist a subsequence $(u'_{n_{k_j}})_{j \ge 1}$ and $g \in L^{p}(I)$ such that $u'_{n_{k_j}} \rightharpoonup g$ weakly in $L^{p}(I)$ if $1 < p < \infty$ or weakly-$\star$ in $L^{\infty}(I)$ if $p=\infty$. But we have
$$
\int_I u_{n_{k_j}} \varphi'
= -\int_I u'_{n_{k_j}} \varphi \quad \forall\varphi \in C^\infty_c(I)
$$
so that in the limit we obtain
$$
\int_I u \varphi'
= -\int_I g \varphi \quad \forall\varphi \in C^\infty_c(I),
$$
which implies $g=u'$ an $u \in W^{1,p}(I)$.
Finally, we have that any subsequence of $(u'_{n_k})_{k \ge 1}$ has a further subsequence that converges to $u'$ (for the weak topology if $1 < p < \infty$ or the weak-$\star$ topology if $p=\infty$), hence the whole sequence $(u'_{n_k})_{k \ge 1}$ converges.


TITLE: Simplifying $a = \dfrac{\sqrt{x}}{x+3} $
QUESTION [0 upvotes]: Solving equations involving terms of the form $ \dfrac{3x}{6x^2} $ is easy. You can cancel the $x$ in the numerator and end up with: $ \dfrac{3}{6x} $.
However, I am presented with an equation of the following form:
$$a = \dfrac{\sqrt{x}}{x+3} $$
Where, $a$ is a constant.
Trying to cancel out the variable $x$ in either the numerator or denominator doesn't work.
How do I go about and solve such an equation?

REPLY [1 votes]: One should be careful with simplifications and with squaring. For instance, the equation $x=1$ has just one solution, but $x^2=1^2$ has two.
Since $x=(\sqrt{x}\,)^2$, you can set $y=\sqrt{x}$ and try to solve
$$
\frac{y}{y^2+3}=a
$$
with the limitation that $y\ge0$ (which holds by definition of square root).
Now, before going on, we observe that there is just the solution $y=0$ (so $x=0$) when $a=0$. There is no solution for $a<0$, because the left-hand side is $\ge0$ for every $y\ge0$.
So we proceed to solve the equation with the additional assumption that $a>0$; in particular $y=0$ is not a solution.
Removing the denominator and reordering we obtain $ay^2-y+3a=0$. We note that the sum of the roots is $3/a>0$ and the product is $3a/a=3>0$. Thus the equation has two positive roots (possibly coincident) as soon as its discriminant is $\ge0$: thus the condition is
$$
\Delta=1-12a^2\ge0
$$
Under this condition you can write the solutions as
$$
y=\dfrac{1-\sqrt{\Delta}}{2a}\qquad\text{or}\qquad y=\dfrac{1+\sqrt{\Delta}}{2a}
$$
and you can square these to find the solutions for $x$.
In conclusion: the equation has no solution for $a<0$; one solution for $a=0$; two solutions for $0<a<1/\sqrt{12}$; one solution (with multiplicity two) for $a=1/\sqrt{12}$; no solution for $a>1/\sqrt{12}$.