TITLE: Geometric interpretation of elliptic curve point addition in projective space
QUESTION [1 upvotes]: I am familiar with the idea of considering an elliptic curve in projective space, particularly thanks to this excellent Crypto SE post. What I find most satisfying about it is the fact that it unifies the usual elliptic curve points $(x,y)$ with the very distinct point $\mathcal{O}$.
In affine space we can interpret addition by drawing a chord between the two points (or tangent if $P_1=P_2$), finding its intersection with the rest of the curve, and reflecting across the $x$-axis. Unfortunately this falls apart for additions involving $\mathcal{O}$.
My question is whether there is an analagous interpretation of addition in projective space which also works for $\mathcal{O}$. I was hoping to discover one by treating chords between points as planes, but haven't been able to come up with anything.

REPLY [2 votes]: Let me try to formulate the group operation in purely projective terms, without any use of coordinates and without a designated line at infinity.
To add $P$ and $Q$, you connect them with a line, and intersect that line with the curve to obtain $R$ as the third point of intersection. Then you connect $R$ to $\mathcal O$ with another line, and obtain $S$ as the third point of intersection. That point is the result of your addition.
In the regular affine picture following established conventions, $\mathcal O$ is $[0:1:0]$, the point at infinity in $y$ direction. So the second line in the description above would be vertical (constant $x$ coordinate) and $S$ will be the mirror image of $R$.
So how does this behave if one of the points of the addition is $\mathcal O$? Suppose w.l.o.g. that $Q=\mathcal O$. So the first line is the line defined by $P$ and $\mathcal O$. The third point on that line is $R$. Then the second line is defined by $R$ and $\mathcal O$, so it is the same as the first. And the third point of intersection is now again $P$, leading to the expected identity $P+\mathcal O=P$. Working as expected.

treating chords between points as planes

Note that the vocabulary I chose above was purely 2d: points, lines and a curve. There is some value in considering a 3d embedding when you first learn about projective plane, homogeneous coordinates and so on. But in my experience that view can become distracting in the long run for most operations. Yes, you can view some coordinate vector as the normal vector of a plane which intersects the drawing plane in the line you are talking about, but that usually doesn't really matter. All that matters is that two points define a line, even if one or both of them are at infinity. Two lines intersect in a point, even if they are parallel or one of them is at infinity. And a line intersects an elliptic curve in three points, or perhaps one if the field is not algebraically closed. If you have a decent intuition for these concepts, imagining this embedded in 3d will add little value but complicate vocabulary.


TITLE: List of complexity classes closed under complement
QUESTION [2 upvotes]: Is there a list of 'natural' complexity classes closed under complement? Some that I could think of are P, ZPP, BPP, NP $\cap$ co-NP, PH, PP and PSPACE but surely there are others. Wikipedia and the Complexity-Zoo don't have such a list as far as I can tell but perhaps there are other resources?

REPLY [6 votes]: Nonderministic Logspace (NL) is closed under complement. That is, $NL=coNL$. This non-trivial result is known as (or rather, is the consequence of) the Immerman–Szelepcsényi theorem.

REPLY [3 votes]: All deterministic classes are (almost trivially) closed under complement:
Given a deterministic TM $M$ for a language $L$, one obtains a deterministic TM for its complement by simply running $M$ and then "flipping" the answer. The machine runs in the same time and space asymptotically. By applying the same proof to the complement class, it follows that $L=co-L$.
The same obviously holds for all classes where the result can be safely "flipped", e.g. $ZPP$. Nondeterministic classes obviously fail this requirement.

REPLY [2 votes]: Shaull already mentioned NL.  The same thing also applies to all super-logarithmic space classes.  The only other examples I can find for which I don't know of as-natural

equivalent definitions from which closedness under complement trivially follows are:

SAC and each of its positive levels
and
SZK
.


TITLE: Is this the only planar cubic graph with six squares and two hexagons?
QUESTION [0 upvotes]: Is this the only planar cubic graph with six squares (five inside, one outside) and two hexagons?

How to prove that?
Allowing for multiple edges I can find another, but without not...

REPLY [1 votes]: For a planar cubic graph $G$, we have $E=\frac{3}{2}V$, where $E$ and $V$ denote the number of edges and vertices of $G$, respectively. The number of faces is $F=8$ by assumption. If $G$ is connected then we have $V-E+F=2$, and hence $V=12$ and $E=18$.
Consider one of the hexagonal faces; it is bounded by six edges connecting six vertices. As $G$ is cubic, each of these vertices has precisely one 'outgoing' edge, i.e. an edge that does not bound this hexagon. As there are no triangular faces, no two of these outgoing edges are adjacent, yielding the following configuration:

A bit of trial and error shows that there is only one way to make this graph into a planar cubic graph without triangles:

See Tony Huynh's answer for a rigorous proof of the latter claim.
This is the graph you already found, and this is the only such graph.


TITLE: Proof the limit of $c_n$ when $c_n=a_n$ for odd n and $c_n=b_n$ for even n
QUESTION [0 upvotes]: Let us consider two sequences $(a_n)$ and $(b_n)$ with equal limits denoted as $g$ for $n \to \infty$. Let us take
$$c_n=\left\{
\begin{array}{ccc}
a_n&\mbox{for odd n,}\\
b_n&\mbox{for even n.}
\end{array}
\right.$$
Using the definition of the limit show that $\lim_{\,n \to \infty} c_n = g$.
Let us take $\epsilon>0.$ Since $\lim_{\,n \to \infty} a_n = g$ and $\lim_{\,n \to \infty} b_n = g$, we have that there is an $N_1 \in \mathbb{N}$ and an $N_2 \in \mathbb{N}$ such that for all $n>N_1$
$$|a_n-g|<\epsilon$$ and for all $n>N_2$ $$|b_n-g|<\epsilon.$$
I do not know how to use this to show that $|c_n-g|$ is also bounded. I would be grateful for any help.

REPLY [3 votes]: Take $N := \max\{N_1,N_2\}$. Then for $n > N$, if $c_n = a_n$ then $|c_n - g| = |a_n - g| < \epsilon$ as $n > N_1$. If $c_n = b_n$ then $|c_n - g| = |b_n - g| < \epsilon$ as $n > N_2$.


TITLE: Graph Theory question about path counting.
QUESTION [1 upvotes]: How many different (open) paths of length $k$; $(0 < k < 2n)$ are there in a complete bipartite graph $K_{n,n}$; $(n>2)$ ? We do not consider different those paths that have the same or reversed sequence of nodes.

REPLY [2 votes]: If $k$ is odd, the path visits $(k+1)/2$ nodes on each side. Without loss of generality we can consider the node on the left as the begining of the path and give it an orientation. The total number of paths is $\frac{n!}{(n-(k+1)/2)!} \frac{n!}{(n-(k+1)/2)!}$
since any possible ordering of $(k+1)/2$ nodes on the left side and $(k+1)/2$ nodes on the right side is valid.
If $k$ is even, the path visits $k/2$ node on one side and $k/2+1$ on the other. Consider first paths that start on the left. The total number of such oriented paths is $\frac{n!}{(n-k/2)!} \frac{n!}{(n-k/2-1)!}$
since any possible ordering of $k/2+1$ nodes on the left side and $k/2$ nodes on the right side is valid. Since we don't care about orientation, we are double counting each path so the total number of unoriented paths that start on the left is $\frac 1 2 \frac{n!}{(n-k/2)!} \frac{n!}{(n-k/2-1)!}$. The same holds for paths that start on the right so in total we have $\frac{n!}{(n-k/2)!} \frac{n!}{(n-k/2-1)!}$ paths.
The answer is therefore for all $k$, $$\frac{n!}{(n-\lceil(k+1)/2\rceil)!} \frac{n!}{(n-\lfloor(k+1)/2\rfloor)!}$$

REPLY [1 votes]: If by "path" we mean a simple (non-self-intersecting) path, then it is relatively easy to count these within a complete bipartite graph $K_{n,n}$.
Let $A = \{a_1,\ldots,a_n\}$ and $B = \{b_1,\ldots,b_n\}$ be the two "parts" of the graph.  Then a path of length $k$ might begin either with a vertex of part $A$ or of part $B$.  If the path is of odd length, then it's reversal will be a path that starts in the opposite part (and we want to count these as a single path).
Odd length paths
Let $k = 2m-1$, and there will be $m$ vertices in part $A$ as well as in part $B$.  All possible sequences in $A$ and $B$ can arise (because the bipartite graph is complete).  So the counting if we start in part $A$ will be:
$$\frac{(n!)^2}{(n-m)!^2}$$
as every such path can be made from an arbitrary distinct $m$-sequence in $A$ interleaved with an arbitrary distinct $m$-sequence in $B$ (and it is unnecessary duplication to consider the paths starting in $B$).  Note that the $2m$ "fence posts" give rise to the $k = 2m-1$ edges that connect them.
Even length paths
Very similar, except that we will have an odd number of vertices, so the path will begin and end in the same "part" $A$ or $B$ of the graph, and we therefore count the two cases separately.  However the sequence in the "dominant" part can be reversed, with an equal number of the shorter distinct sequences in the "lesser" part, so we divide by two in each case.
The two halves of each case (starting in $A$, starting in $B$) make a symmetric whole, so again it suffices to count only the cases starting in $A$.  If $k=2m$, we then have $m+1$ vertices in $A$ to pick and $m$ vertices in $B$, so the count is:
$$ \frac{(n!)^2}{(n-m)!(n-m-1)!} $$


TITLE: How do you calculate the modulo of a really high number with a large power, with a really high mod number?
QUESTION [4 upvotes]: I need to work out $516489222^{22} \pmod{96899}$. I know there are easier ways of working this out, but am really struggling.

REPLY [7 votes]: Here's a somewhat high-level description of the several tricks one can use to make numbers smaller and computations "easier". You may need some knowledge of elementary number theory to understand these tricks.
Things can be made easier by first factoring $96899 = 11 \cdot 23 \cdot 383$. Then, to find $$516489222^{22} \pmod{96899},$$ it suffices to find
$$\begin{align} &516489222^{22} \pmod{11}, \tag{1.} \\ &516489222^{22} \pmod{23}, \tag{2.} \\ &516489222^{22} \pmod{383}, \tag{3.} \end{align}$$
and then combine the results using the Chinese Remainder Theorem. For each of these three cases, we will use the fact that for all $a \not\equiv 0 \pmod{m}$, we have
$$a^b \equiv (a \mod m)^{(b \mod \phi(m))} \pmod m,$$
to reduce the numbers in the computations considerably. Here $\phi$ is Euler's totient function, and being able to reduce exponents with multiples of $\phi(m)$ is a consequence of Euler's theorem. (In our applications $m$ is always prime, in which case the theorem is also known as Fermat's little theorem, where $\phi(p) = p-1$.)

For the first, using that $516489222 \equiv -4 \pmod{11}$ and $22 \equiv 2 \pmod{\phi(11)}$,
we get
$$516489222^{22} \equiv (-4)^2 \equiv 16 \equiv 5 \pmod{11}.$$
For the second, using that $516489222 \equiv 3 \ [\not\equiv 0] \pmod{23}$ and $22 \equiv 0 \pmod{\phi(23)}$, we get
$$516489222^{22} \equiv 3^0 \equiv 1 \pmod{23}.$$
For the last, using that $516489222 \equiv -66 \pmod{383}$, we get
$$516489222^{22} \equiv (-66)^{22} \pmod{383}.$$
We do not get around doing some modular computations here, but the work can be reduced a bit by using square-and-multiply:
$$516489222^{22} \equiv (-66)^{22} \equiv ((-66)^2)^{11} \equiv 143^{11} \equiv 143 \cdot (143^2)^5 \equiv 143 \cdot 150^5 \\ \equiv 143 \cdot 150 \cdot (150^2)^2 \equiv 2 \cdot (-97)^2 \equiv 2 \cdot 217 \equiv 51 \pmod{383}.$$

So concluding, we have
$$516489222^{22} \equiv \begin{cases} 5 \pmod{11} \\ 1 \pmod{23} \\ 51 \pmod{383} \end{cases}$$
Combining them using the CRT, we get
$$516489222^{22} \equiv 4647 \pmod{11 \cdot 23 \cdot 383 = 96899}$$

REPLY [2 votes]: The Chinese Remainder Theorem can be helpful in this situation. Since
$96899=11*23*383$ it suffices to calculate the number $\pmod{11}, \pmod{23}$ and $\pmod{383}$.
$516489222 \equiv 7 \pmod{11}$. By Fermat Little Theorem
$$(516489222)^{22} \equiv ((516489222)^{10})^27^2 \equiv 1 \cdot 7^2 \equiv 5 \pmod {11}$$
$516489222 \equiv 3 \pmod{23}$. By Fermat Little Theorem
$$(516489222)^{22} \equiv  1 \pmod {23}$$
$516489222 \equiv 317 \equiv -66 \pmod{383}$. Then
$$(516489222)^{22} \equiv (-66)^{22} \equiv  (143)^{11} \pmod {383}$$
Now
$$(143)^{2} \equiv 150 \pmod {383}$$
$$(143)^{4} \equiv 150^2 \equiv -97 \pmod {383}$$
$$(143)^{8} \equiv (-97)^2 \equiv 217 \pmod {383}$$
$$(143)^{11} \equiv (143)^{8} (143)^{2} 143 \equiv 217 \cdot 150 \cdot 143 \equiv -5 \cdot 143 \equiv 51 \pmod {383}$$
Now, you can use the Chinese Remainder Theorem to find the unique class $\pmod{96899}$ which is $7 \pmod{11}, 1 \pmod{23}$ and $51 \pmod{383}$


TITLE: Is this a correct Monte-Carlo expression for $\pi$?
QUESTION [10 upvotes]: I have a bicycle with one of those O-locks on it and too often when I park the bike and I want to lock it, the lock hits one of the spokes of the rim. This can be frustrating and surprises me that it occurs so often. I mean, the spokes are so thin and not that many really so one would think that this should not happen that often (like every day or so). And so every time this happened I reminded myself to calculate the probability of this happening. I just did it (after a year) and now I want to know what you guys think about my calculation, is it OK?
This is not a textbook example so there is no answer to look up or anything, that's why I need your feedback.
I have modeled the situation as shown in the figure below, where I have included one spoke only.
Let the O-lock have diameter $d_{L}$ and the spoke have a diameter $d_S$. Let $R$ denote the "radius" from center of wheel to the point where the O-lock comes and goes (this is approximately equal to the radius of the rim). Then, we have that the corresponding angles are given by $w_S=d_S/R$ and $w_L = d_L/R$ so that the spoke will hit the lock (the shaded disk in figure below) when $\theta$ is in the interval $[0, w_L+w_S]$ modulo $2\pi$ (See figure: $R$ is fixed so the point $(R,\theta)$ is on a circle). Then the probability for hit (wheel with one spoke) is given by
$$ P(1) = \frac{w_L+w_S}{2\pi}\times 1 = \frac{1}{2\pi R}(d_L+d_S).$$
Generalizing to $N$ equally spaced spokes we get
$$ P(N) =\begin{cases} \frac{N}{2\pi R}(d_L+d_S)& \text{for } N=0,1\dots\\1 & \text{for } N\geq M= \Big\lceil\frac{2\pi R}{d_L+d_S}\Big\rceil\end{cases}. $$
Example: plugging in some typical numbers; $N=36, R\approx 0.3$m, $d_S\approx 2\times 10^{-3}$m , $d_L\approx \times 10^{-2}$m we find
$$P(36) \approx 0.23$$ which kind of agrees with my everyday experience of this problem.
I guess one could obtain a (Monte-Carlo)-value for $\pi$ this way?
Could someone tell me what I have done wrong above? Or perhaps derive the correct expression for the probability?

REPLY [1 votes]: A simpler method is as follows:
Find out the diameter of the cross section of your lock. Now calculate the area of the tyre in which the lock locks the tyre. (this can be done by the formula:
         $π(D^2-d^2)$,with d being the distance from the center of the rim to the lower end of the lock's cross section and $D$ being the length
 between the centre and the upper end of the cross section)
now find out the area of all the spokes lying in that region by $(D-d)x$(thickness of spokes) and subtract with the area found earlier. Find the area of the cross section of the lock and you can get the probabilty by dividing it by the figure found earlier. You can also find the anti probability too. Suggestions welcome


TITLE: How this formula return intersection between circle and line?
QUESTION [0 upvotes]: circle line intersection
as you can see in the picture, the writer said
the horizontal = radius * sin(theta)
the vertical = 1 - radius * cos(theta)
i know x = rcos(theta), y = rsin(theta) but I can't understand where this formula came from.
and the method is
private bool SimulateSphereCast(Vector3 groundNormal, out RaycastHit hit)
        {
            float groundAngle = Vector3.Angle(groundNormal, controller.up) * Mathf.Deg2Rad;

            Vector3 secondaryOrigin = controller.transform.position + controller.up * Tolerance;

            if (!Mathf.Approximately(groundAngle, 0))
            {
                float horizontal = Mathf.Sin(groundAngle) * controller.radius;
                float vertical = (1.0f - Mathf.Cos(groundAngle)) * controller.radius;

                // Retrieve a vector pointing up the slope
                Vector3 r2 = Vector3.Cross(groundNormal, controller.down);
                Vector3 v2 = -Vector3.Cross(r2, groundNormal);

                secondaryOrigin += Math3d.ProjectVectorOnPlane(controller.up, v2).normalized * horizontal + controller.up * vertical;
            }

            if (Physics.Raycast(secondaryOrigin, controller.down, out hit, Mathf.Infinity, walkable, triggerInteraction))
            {
                // Remove the tolerance from the distance travelled
                hit.distance -= Tolerance + TinyTolerance;

                return true;
            }
            else
            {
                return false;
            }
        }
    }

And whole articles is here

REPLY [0 votes]: The formula in the figure is wrong.
It should say $(1 - \cos (\theta)) \times \mathop{radius}.$
It is $1 - \cos(\theta)$ because the distance is measured from the bottom of the circle, not from the center.
The vertical distance of that point from the center of the circle is
indeed $\cos(\theta)\times \mathop{radius}.$
The vertical distance from the center to the bottom of the circle is
$\mathop{radius}.$
The vertical distance shown in the figure is the remaining vertical distance to the bottom after you have already come down a distance $\cos(\theta)\times \mathop{radius}$ from the center of the circle.
This remaining distance is
$$ \mathop{radius} - \cos (\theta) \times \mathop{radius}
=(1 - \cos (\theta)) \times \mathop{radius}.$$


TITLE: Characteristic polynomial of square of matrix?
QUESTION [2 upvotes]: Let $A \in M_n(\mathbb{C})$ with $p_A(x)=(-1)^n\cdot (x-λ_1)\dots(x-λ_n)$ it's characteristic polynomial.
Prove that $p_{A^2}(x)=(-1)^n\cdot (x-λ_1^2)\dots(x-λ_n^2)$
Any tips on where to start from?

REPLY [2 votes]: The characteristic polynomial of a matrix has roots that are equal to its eigenvalues. Let $x$ be an eigenvector of $A$ with corresponding eigenvalue $\lambda$.  Then
$$
AAx = A \lambda x = \lambda^2 x.
$$
So $x$ is an eigenvector of $A^2$ too, but its eigenvalue is $\lambda^2$.  So the roots of the characteristic polynomial of $AA$ are the squares of the roots of $A$.  Its characteristic polynomial is completely determined by this fact.


TITLE: Continuous bounded function $f:\mathbb{R}\rightarrow \mathbb{R}$
QUESTION [6 upvotes]: Question is to check which of the following holds (only one option is correct) for a continuous bounded function  $f:\mathbb{R}\rightarrow \mathbb{R}$.

$f$ has to be uniformly continuous.
there exists a $x\in \mathbb{R}$ such that $f(x)=x$.
$f$ can not be increasing.
$\lim_{x\rightarrow \infty}f(x)$ exists.

What all i have done is :

$f(x)=\sin(x^3)$ is a continuous function which is bounded by $1$ which is not uniformly continuous.
suppose $f$ is bounded by $M>0$ then restrict $f: [-M,M]\rightarrow [-M,M]$ this function is bounded ad continuous so has fixed point.
I could not say much about the third option "$f$ can not be increasing". I think this is also true as for an increasing function $f$ can not be bounded but i am not sure.
I also believe that $\lim_{x\rightarrow \infty}f(x)$ exists as $f$ is bounded it should have limit at infinity.But then I feel the function can be so fluctuating so limit need not exists. I am not so sure.

So, I am sure second option is correct and fourth option may probably wrong but i am not so sure about third option.
Please help me to clear this.
Thank You. :)

REPLY [5 votes]: $\tan^{-1}x$ is increasing. $\sin (x^3)$ has no limit at infinity.


TITLE: Finding the Convergence Domain of the Taylor series of $\sin (z_1 + z_2^2)$ at the origin
QUESTION [1 upvotes]: I have to find the convergence domain $\mathcal{D}$ of the  Taylor series of $f(z) =\sin (z_1 + z_2^2)$ at the origin. I claim $\mathcal{D}=\mathbb{C}^2$ and I prove this in the following manner:
We have $$\sin (z_1 + z_2^2) = \sin(z_1)\cos(z_2^2) + \sin(z_2^2)\cos(z_1)$$
Since the taylor series of each multiplicand of each summand converge everywhere (as a single variable complex function), so does $f(z)$, therefore the claim is proved.
Is my proof/argument correct?

REPLY [1 votes]: Just like in one variable, the Taylor series around $p$ will converge on any polydisc centered at $p$ that fits in the domain on which the function is holomorphic.  The way to prove it is just the iterated Cauchy integral formula leading to the several variable Cauchy estimates.  See e.g. https://www.jirka.org/scv/scv.pdf sections 1.1 and 1.2, and in particular Theorem 1.2.1.  Since your function is holomorphic in ${\mathbb C}^2$ and so any polydisc fits in your domain, and so the series centered at any point will converge in the entire ${\mathbb C}^2$.  Your argument would also work as long as you can use that the product and addition of power series has at least that same domain of convergence, which is not as easy to prove from scratch, but it is elementary.


TITLE: Number of eigen vectors
QUESTION [0 upvotes]: Assume we have square matrix $A_{N \times N}$.
Can we say that the maximum number of independent eigen vectors, with nonzero eigenvalues, equals $\text{Rank}(A)$?
We can say that such number must be smaller equal than $\text{Rank}(A)$, because an eigenvector with nonzero eigenvalue means that it must belongs to $\text{Range}(A)$, so the independent set of such vectors must be a basis of some subspace of $\text{Range}(A)$, thus not exceeding $\text{dim} \{  \text{Range}(A) \} = \text{Rank} (A)$.
But my question is, does the equality holds?

REPLY [0 votes]: No. Let $A=\begin{bmatrix}1&1\\0&1\end{bmatrix}$. $1$ is the only eigenvalue of $A$ and $ \dim ker(A-I)=1$. But $rank(A)=2$


TITLE: A complex integral (explicit solution?)
QUESTION [0 upvotes]: Let $r<1$ be a positive number close to 1. How can I compute
$$
\int_0^{2\pi}\frac 1 {e^{i\theta}-r}d\theta?
$$
Can this be computed in a closed form? Thanks!

REPLY [1 votes]: The answer is $0$. By Cauchy's Theorem $\int_C \frac 1 {1-rz}=0$. This gives $\int_0^{2\pi} \frac {ie^{i\theta}} {1-re^{i}\theta} d\theta=0$. Hence  $\int_0^{2\pi} \frac 1 {e^{-i\theta}-r} d\theta=0$. Can you take it from here?


TITLE: Sum of the geometric series $ \frac{1}{(2^{k})^2}+...+\frac{1}{2^2}+1 $
QUESTION [0 upvotes]: Did I get the sum of this geometric sequence right? (Last few terms fix is geometric now)
$$ \frac{1}{(2^{k})^2}+\frac{1}{(2^{(k-1)})^2}+\dots+\frac{1}{(2^2)^2}+\frac{1}{(2)^2}+1 $$
Using

$S_n = \frac{a_1(r^n-1)}{r-1}$ with $a_1 =1,r=\frac{1}{2^2}, n=k \Rightarrow S_n = \frac{4-4^{1-k}}{3}$

REPLY [2 votes]: A sum such as $$\frac1{2^{k^2}} +\frac1{2^{(k-1)^2}} +\frac1{2^{(k-2)^2}} +\frac1{2^{(k-3)^2}} +\ldots +\frac1{2^2} +1$$ for positive $k$ isn't a geometric series because the exponents don't follow an arithmetic progression. Also, how come the second-to-last term is $1/2^2?$


TITLE: Zero gravity area
QUESTION [1 upvotes]: How can I find the radius of the circle (average) centered at the center of the earth where the gravitational force of the earth cancels the gravitational force of the other celestial bodies?
Will the velocity which is required to reach the circumference of the circle be the escape velocity of earth? I think because the net acceleration on the object outside the circle will be directed away from the earth.. so that the body has truly escaped the earths gravity..?
Sorry for my non scientific language im not even a layman.

REPLY [1 votes]: Your question is a bit vague. There are a lot of objects moving around the solar system, and they are not always in the same place.
That said, there are two objects that affect the Earth in a more or less constant way: the Moon and the Sun.
The gravity of two massive bodies rotating each other, provided that one of them is quite bigger than the other (such as the Earth-Moon or the Sun-Earth systems), cancel at several points: those are the Lagrangian points.
The nearest one, the Earth-Moon L1 is about 326000 km high.
About your question:

Will the velocity which is required to reach the circumference of the circle be the escape velocity of earth?

No, the escape velocity of Earth is (by definition) the velocity needed to reach an infinite distance from the surface of the Earth. To reach the Earth-Moon L1 you need less speed because:

It is nearer than infinity (obviously).
You have help from the gravity of the Moon.

That said, there are another point where the Earth and Moon gravity cancel out, deep inside the Earth. Since the gravity as you go deep inside the planet decreases (proportional to the distance to the center) while the Moon gravity changes little (350000 km vs 356000 km is not a big deal), there is a point half-way (I didn't do the calculations) where the gravity of both the moon and the Earth would cancel out.


TITLE: Why covariant derivative is partial derivative in SR?
QUESTION [0 upvotes]: I'm new in these topics and i've been confuse at some relations between the limit of SR for GR.
In cartesian coordinates, basis do not change, so
\begin{equation}\Gamma^{\mu}_{\alpha\beta}=0 \quad \quad(1)\end{equation}
and thus $V^{\alpha};_{\beta}=V^{\alpha},_{\beta}$.
I've seen that polar coordinates changes point to point:
\begin{equation}|\vec{e_{\theta}}|^{2}=r^{2}\quad\quad(1.1),\end{equation} so polar coordinates has non constant basis and for these cordinates, equation (1) isn't true, as same than the covariant derivative isn't the partial derivative for that cordinate system, that is
\begin{equation}V^{\alpha};\beta\ne V^{\alpha},\beta\quad (1.2)\end{equation}
My book Schutz define a locally inertial frame as a frame in which everything is locally like SR, and formally by a theorem we see that

Choose any point $\mathcal{P}$ of the manifold, a coordinate system can be found whose origin is at $\mathcal{P}$ and which the metric is Minkowski metric (2)

He also says

In particular, we say that the derivatives of the basis vectors of a locally inertial coordinate system are zero at $\mathcal{P}$' (3)

Why ???, the coordinate system is arbitrary, can't be polar or spherical ?, i understand that locally the inertial frame has behaviour as SR but in SR i can have polar, spherical and some other's coordinate system's that derivative of basis are not equals to zero.
This definition mentioned in (3) leads he to conclude that
\begin{equation} V^{\alpha};\beta=V^{\alpha},\beta \quad \mbox{      at } \mathcal{P} \mbox{ in this frame } \quad \quad (4)\end{equation}
Why ?? again this frame is in SR, okay, but this frame can assume a lot of coordinate systems that derivative of basis isn't necessary equals to zero, but if i consider this point $\mathcal{P}$ as a instantaneous measure, the derivative of non constant basis will be zero, but how $V^{\alpha},\beta$ exists in instantaneous measure ??
than he follows with these conclusions
\begin{equation}g_{\alpha\beta;\gamma}=g_{\alpha\beta,\gamma}=0\quad \mbox{ at } \mathcal{P}\quad \quad (5)\end{equation}
because the last equation is a tensor equation, so is valid

\begin{equation}g_{\alpha\beta;\gamma}=0\quad \mbox{in any basis }\quad (6) \end{equation}

this will be valid in any basis in this SR frame ?? at SR frame near point $\mathcal{P}$, or in all frames ??.
All these doubts lead me to ask myself if is SR defined only for Cartesian coordinates, that is, when we talk about SR is supposed to assume only Cartesian coordinates ??.
Sorry about a ton of question, i'm lost in these doubts a few days ago

REPLY [2 votes]: Why ???, the coordinate system is arbitrary, can't be polar or spherical ?, i understand that locally the inertial frame has behaviour as SR but in SR i can have polar, spherical and some other's coordinate system's that derivative of basis are not equals to zero.

All of those statements of Schultz revolve around the idea of a coordinate system whose origin is $\mathcal P$, and in which $\Gamma^i_{jk}(\mathcal P)=0$;  a polar coordinate chart (of a smooth manifold) cannot even include the coordinate origin, so such coordinate systems are not suitable for this kind of analysis.
I admit, Schultz is perhaps not so good at making this distinction.  What you say is essentially true - polar coordinates are as good as any other coordinates - with the exception that they are poorly behaved at the origin.


TITLE: Descend finite etale algebras
QUESTION [2 upvotes]: Let $\pi:X\to\mathcal X$ be a presentation of an Artin stack $\mathcal X$ of finite type over a field $k,$ and let $f:Y\to X$ be a finite \'etale covering. Does there exist a finite \'etale covering $Y'\to X$ factoring through $Y,$ such that $Y'$ can be given descent structure, i.e. there exists an isomorphism $pr_1^*Y'\cong pr_2^*Y'$ over $X\times_{\mathcal X}X$ satisfying cocycle condition, so that $Y'\to X$ descend to a finite \'etale covering $\mathcal Y\to\mathcal X$?

REPLY [4 votes]: I don't think so (finite etale covers cannot be localized in smooth topology in the sense that you describe). Say, $\mathcal{X}$ is a point, and $X$ is a smooth variety with non-trivial fundamental group, say, an elliptic curve (or $\mathbb{A}^1-\{0\}$). Then $\pi$ is a presentation. Let $f:Y\to X$ be a non-trivial finite etale cover, say, the cover of the elliptic curve by an isogeneous elliptic curve. Then your question becomes: `is there a trivial (i.e., lifted from $\mathcal{X}$) cover $Y'$ of $X$ with a map to $Y$? This is of course not true.


text:
\begin{document}

\nocite{*}

\maketitle

\begin{abstract} We study the theta decomposition of Jacobi forms of nonintegral lattice index for a representation that arises in the theory of Weil representations associated to even lattices, and suggest possible applications. \end{abstract}

\section{Introduction}

This note is concerned with modular forms for the (dual) Weil representation $\rho^*$ attached to an even lattice $\Lambda$ with quadratic form $Q$. Letting $A = \Lambda'/\Lambda$ denote the discriminant group of $\Lambda$, this is a representation of the metaplectic group $Mp_2(\mathbb{Z})$ on $\mathbb{C}[A]$. Weil representations are a family of representations with better-than-usual arithmetic properties; for one thing, there are bases of modular forms in every weight with integral coefficients \cite{McG}. (Inducing the trivial representation from noncongruence subgroups to $SL_2(\mathbb{Z})$ gives plenty of examples of representations for which this fails.) These representations are important in the context of theta lifts such as \cite{B}. \par \

When $\Lambda$ is positive-definite it is well-known that modular forms for $\rho$ are essentially the same as Jacobi forms of lattice index $\Lambda$. This is a higher-dimensional version of the theta decomposition of chapter 5 of \cite{EZ}. One immediate advantage of Jacobi forms is their natural algebraic structure; multiplying Jacobi forms corresponds to an unintuitive operation on vector-valued modular forms. \par \

In the vector-valued setting, many objects that are usually indexed by integers $n \in \mathbb{Z}$ have to be replaced by pairs $(n,\gamma)$ with $\gamma \in A$ and $n \in \mathbb{Z} - Q(\gamma)$ (as in the final paragraph of the introduction of \cite{Br}.) In particular we might consider Jacobi forms of index $(m,\beta)$ where $\beta \in A$ and $m \in \mathbb{Z} - Q(\beta)$ is not generally integral. Such Jacobi forms of rational index have appeared in \cite{W}. \par \

The general theory of Jacobi forms indexed by a rational, symmetric matrix was worked out by Raum \cite{R}. Nonintegral indices force the Heisenberg group to act through a nontrivial representation. The irreducible representations that can arise were classified in \cite{R}, where theta decompositions that reduce all such Jacobi forms to vector-valued modular forms also appear. \par \

In the first half of this note we define a family of (generally not irreducible) representations $\sigma_B^*$ of the Heisenberg group that appear when one considers the Fourier-Jacobi expansion of Siegel theta functions. For these Jacobi forms the theta decomposition is particularly explicit and has a simpler proof; it is little more than applying an index-raising Hecke operator to reduce to the theta decomposition of integral-index Jacobi forms. \par \

In the second half of this note we suggest some applications. We remark that Eisenstein series and Poincar\'e series match up under this isomorphism (proposition 8) so computing bases of rather general Jacobi forms reduces to Bruinier's formulas (\cite{Br}, section 1.2). We point out that setting $z=0$ in the resulting Jacobi forms leads to some isomorphisms between modular forms for lattices of unrelated discriminants and work out an amusing example (example 11). Finally we give another interpretation of quasi-pullbacks of Borcherds products (as in \cite{M}) and construct some Borcherds products of small weight on lattices of signature $(4,2)$. \par

\section{Jacobi forms of lattice index}

For simplicity we will work with fixed Gram matrices, rather than abstract discriminant forms or even lattices. Therefore let $\mathbf{S}$ be a Gram matrix (that is, a nondegenerate, symmetric integer matrix with even diagonal) of size $e \in \mathbb{N}$ and let $\rho^* = \rho^*_{\mathbf{S}}$ denote the (dual) Weil representation of the lattice $\mathbb{Z}^{e}$ with quadratic form $Q(x) = \frac{1}{2}x^T \mathbf{S}x$ and bilinear form $\langle x,y \rangle = Q(x+y) - Q(x) - Q(y)$ (as in \cite{B},\cite{Br}). If $A = A(\mathbf{S})$ denotes the discriminant group $\mathbf{S}^{-1} \mathbb{Z}^{e} / \mathbb{Z}^{e}$ and $\mathfrak{e}_{\gamma}$, $\gamma \in A$ denotes the natural basis of the group ring $\mathbb{C}[A]$ then on the usual generators $$S = \Big( \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \sqrt{\tau} \Big), \quad T = \Big( \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, 1 \Big)$$ of $Mp_2(\mathbb{Z})$, $\rho^*$ is given by \begin{align*} &\rho^*(S)\mathfrak{e}_{\gamma} = \frac{\mathbf{e}(\mathrm{sig}(\mathbf{S}) / 8)}{\sqrt{|A|}} \sum_{\beta \in A} \mathbf{e}\Big( \langle \gamma, \beta \rangle \Big) \mathfrak{e}_{\beta}, \\ &\rho^*(T) \mathfrak{e}_{\gamma} = \mathbf{e}\Big( - Q(\gamma) \Big) \mathfrak{e}_{\gamma}.\end{align*} Here $\mathbf{e}(\tau) = e^{2\pi i \tau}$ and $\mathrm{sig}(\mathbf{S}) = b^+ - b^-$ if $b^+,b^-$ are the numbers of positive and negative eigenvalues of $\mathbf{S}$. \par \

For $N \in \mathbb{N}$, let $\mathcal{H}_{N}$ denote the Heisenberg group of degree $N$: $$\mathcal{H}_{N} = \Big\{ (\lambda,\mu,t) \in \mathbb{Z}^{N} \times \mathbb{Z}^{N} \times \mathbb{Z}^{N \times N}: \; t - \lambda \mu^T \; \text{symmetric} \Big\}$$ with group operation $$(\lambda_1,\mu_1,t_1) \cdot (\lambda_2,\mu_2,t_2) = (\lambda_1 + \lambda_2,\mu_1 + \mu_2, t_1 + t_2 + \lambda_1\mu_2^T - \mu_1 \lambda_2^T).$$ For any matrix $B \in \mathbb{Q}^{e \times N}$ for which $\mathbf{S}B$ is integral, we can define a unitary representation $\sigma_B^*$ of $\mathcal{H}_{N}$ on $\mathbb{C}[A]$ by $$\sigma_B^*(\lambda,\mu,t) \mathfrak{e}_{\gamma} = \mathbf{e}\Big( -\gamma^T \mathbf{S} B \mu - \frac{1}{2} \mathrm{tr}(B^T \mathbf{S} B (t - \lambda \mu^T)) \Big) \mathfrak{e}_{\gamma - B \lambda},$$ and this depends only on $B$ mod $\mathbb{Z}^{e \times N}$. (When $N = 1$ this reduces to the representations $\sigma_{\beta}^*$ of \cite{W}.) These representations are compatible with $\rho^*$ in the sense that $$\rho^*(M)^{-1} \sigma_B^*(\zeta) \rho^*(M) = \sigma_B^*(\zeta \cdot M),$$ where $Mp_2(\mathbb{Z})$ acts on $\mathcal{H}_{N}$ from the right by $$(\lambda,\mu,t) \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} = (a \lambda + c \mu, b \lambda + d \mu, t),$$ so they combine to a representation $\rho_B^*$ of the degree $N$ Jacobi group: $$\rho_B^* : \mathcal{J}_{N} = \mathcal{H}_{N} \rtimes Mp_2(\mathbb{Z}) \longrightarrow \mathrm{GL}\, \mathbb{C}[A],$$ $$\rho_B^*(\lambda,\mu,t,M) = \rho^*(M) \sigma_B^*(\lambda,\mu,t).$$ All of this can be shown directly but it is easier to prove by restricting the Weil representation of Siegel modular groups to their parabolic subgroups. \par \

\begin{defn} A \textbf{modular form} of weight $k \in \frac{1}{2}\mathbb{N}_0$ for the Gram matrix $\mathbf{S}$ is a holomorphic function $f : \mathbb{H} \rightarrow \mathbb{C}[A]$ that satisfies the functional equations $$f\Big( \frac{a \tau + b}{c \tau + d} \Big) = (c \tau + d)^k \rho^*(M) f(\tau), \quad M = \Big( \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \sqrt{c\tau + d} \Big) \in Mp_2(\mathbb{Z})$$ and is bounded in $\infty$. The space of modular forms of weight $k$ will be denoted $M_k(\mathbf{S})$.
\end{defn}

Any modular form can be expanded as a Fourier series $f(\tau) = \sum_{\gamma \in A(\mathbf{S})} \sum_{n \in \mathbb{Z} - Q(\gamma)} c(n,\gamma) q^n \mathfrak{e}_{\gamma}$ where $c(n,\gamma) = 0$ unless $n \ge 0$. It follows from the functional equation under $Z = (-I,i)$ that $M_k(\mathbf{S}) = 0$ unless $k + \mathrm{sig}(\mathbf{S})/2$ is an integer. Moreover the coefficients $c(n,\gamma)$ satisfy $$c(n,-\gamma) = (-1)^{k + \mathrm{sig}(\mathbf{S})/2} c(n,\gamma)$$ so we refer to $k$ as a \textbf{symmetric} or \textbf{antisymmetric weight} as $k + \mathrm{sig}(\mathbf{S})/2$ is even or odd, respectively. \par \

\begin{defn} Let $\mathcal{M} \in \mathbb{Q}^{N \times N}$ be a symmetric positive-definite matrix and let $B$ be a matrix as above. A \textbf{Jacobi form} of weight $k \in \frac{1}{2} \mathbb{N}_0$ and index $(\mathcal{M},B)$ for the Gram matrix $\mathbf{S}$ is a holomorphic function $\Phi : \mathbb{H} \times \mathbb{C}^{N} \rightarrow \mathbb{C}[A]$ that satisfies the functional equations $$\Phi \Big( \frac{a \tau + b}{c \tau + d}, \frac{z}{c \tau + d} \Big) = (c \tau + d)^k \mathbf{e}\Big( \frac{c}{c \tau + d} z^T \mathcal{M} z \Big) \rho^*(M) \Phi(\tau,z), \quad M = \Big( \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \sqrt{c \tau + d} \Big) \in Mp_2(\mathbb{Z})$$ and $$\Phi(\tau, z + \lambda \tau + \mu) = \mathbf{e}\Big(-\tau \lambda^T \mathcal{M} \lambda - \mathrm{tr}\, \mathcal{M}(2 \lambda z^T + t + \mu \lambda^T) \Big) \sigma_B^*(\lambda,\mu,t) \Phi(\tau,z)$$ for all $(\lambda,\mu,t)$ in the Heisenberg group (in particular, the left side of the above equation should be independent of $t$) and for which the usual vanishing condition on Fourier coefficients holds (see below). We denote the space of Jacobi forms as above by $J_{k,\mathcal{M},B}(\mathbf{S})$.
\end{defn}

This transformation law is often expressed through the slash operator $|_{k,\mathcal{M},B}$: $$\Phi \Big|_{k,\mathcal{M},B} M(\tau,z) = (c \tau + d)^{-k} \mathbf{e}\Big( -\frac{c}{c \tau + d} z^T \mathcal{M} z \Big) \rho^*(M)^{-1} \Phi(M \cdot \tau,z)$$ and $$\Phi \Big|_{k,\mathcal{M},B} (\lambda,\mu,t)(\tau,z) = \mathbf{e}\Big( \tau \lambda^T \mathcal{M} \lambda + \mathrm{tr}\, \mathcal{M}(2 \lambda z^T + t + \mu \lambda^T) \Big) \sigma_B^*(\lambda,\mu,t)^{-1} \Phi(\tau,z + \lambda \tau + \mu).$$

Jacobi forms have Fourier expansions in both variables; we write this out as $$\Phi(\tau,z) = \sum_{\gamma \in A(\mathbf{S})} \sum_{n \in \mathbb{Q}} \sum_{r \in \mathbb{Q}^{N}} c(n,r,\gamma) q^n \zeta^r \mathfrak{e}_{\gamma},$$ where $q^n = e^{2\pi i n \tau}$ and $\zeta^r = e^{2\pi i r^T z}$. The vanishing condition referred to above is then $c(n,r,\gamma) = 0$ whenever $4n - r^T \mathcal{M}^{-1} r < 0$. The indices $(n,r,\gamma)$ and $(\mathcal{M},B)$ that can actually occur are also restricted as follows: \par \

\noindent (i) the transformation under $T \in Mp_2(\mathbb{Z})$ implies $c(n,r,\gamma) = 0$ unless $n \in \mathbb{Z} - Q(\gamma)$; \\ (ii) the transformation under $(0,0,e_i e_j^T + e_j e_i^T) \in \mathcal{H}_{N}$ implies $c(n,r,\gamma) = 0$ unless $\mathcal{M} + \frac{1}{2}B^T \mathbf{S}B$ is half-integral, with integral diagonal; \\ (iii) the transformation under $(0,\mu,0) \in \mathcal{H}_{N}$ implies $c(n,r,\gamma) = 0$ unless $r \in \mathbb{Z}^{N} - B^T \mathbf{S} \gamma$. \par \

Moreover, the familiar restriction on weights holds: $J_{k,\mathcal{M},B}(\mathbf{S}) = 0$ unless $k + \mathrm{sig}(\mathbf{S})/2$ is integral, in which case $c(n,-r,-\gamma) = (-1)^{k + \mathrm{sig}(\mathbf{S})/2} c(n,r,\gamma)$ for all Jacobi forms $\sum c(n,r,\gamma) q^n \zeta^r \mathfrak{e}_{\gamma}$. \par \

\begin{rem} Vector-valued modular forms and Jacobi forms for a fixed Gram matrix $\mathbf{S}$ do not have natural ring structures although they are modules over the ring of scalar-valued modular forms $M_*$ in a natural way. However we will make reference to the operation of tensoring at several points. If $\mathbf{S}_1$ and $\mathbf{S}_2$ are two Gram matrices (possibly of different sizes) then tensoring their Weil representations gives $$\rho^*_{\mathbf{S}_1} \otimes \rho^*_{\mathbf{S}_2} \cong \rho^*_{\mathbf{S}_1 \oplus \mathbf{S}_2}, \quad \mathbf{S}_1 \oplus \mathbf{S}_2 = \begin{pmatrix} \mathbf{S}_1 & 0 \\ 0 & \mathbf{S}_2 \end{pmatrix}$$ and therefore the tensor product defines a bilinear map $$\otimes : M_{k_1}(\mathbf{S}_1) \times M_{k_2}(\mathbf{S}_2) \longrightarrow M_{k_1 + k_2}(\mathbf{S}_1 \oplus \mathbf{S}_2)$$ for all valid weights $k_1,k_2$. Similarly if $(\mathcal{M}_1,B_1)$ and $(\mathcal{M}_2,B_2)$ are valid indices for Jacobi forms with the same number of elliptic variables (i.e. the same $N$) then it follows from $$\sigma_{B_1}^* \otimes \sigma_{B_2}^* \cong \sigma_B^*, \quad B = \begin{pmatrix} B_1 \\ B_2 \end{pmatrix}$$ that the tensor product defines a bilinear map $$\otimes : J_{k_1,\mathcal{M}_1,B_1}(\mathbf{S}_1) \times J_{k_2,\mathcal{M}_2,B_2}(\mathbf{S}_2) \longrightarrow J_{k_1+k_2,\mathcal{M}_1+\mathcal{M}_2,B}(\mathbf{S}_1 \oplus \mathbf{S}_2).$$
\end{rem}

\begin{rem} We recover the familiar definition of Jacobi forms of lattice index (e.g. \cite{G}) by restricting to the case that $B=0$ and $\mathrm{det}(\mathbf{S}) = 1$ and by letting $2\mathcal{M}$ be the Gram matrix of the index lattice with respect to any basis.
\end{rem}

\section{Hecke $U$-operators}

The Hecke operators $U_{\ell}$, $\ell \in \mathbb{N}$ of Eichler and Zagier \cite{EZ} generalize immediately to Jacobi forms of lattice index: we define $$U_{\ell} : J_{k,\mathcal{M},B}(\mathbf{S}) \rightarrow J_{k,\ell^2 \mathcal{M}, \ell B}(\mathbf{S}), \quad U_{\ell}\Phi(\tau,z) = \Phi(\tau,\ell z).$$ The map $U_{\ell}$ is injective and $M_*$-linear and its image of $U_{\ell}$ consists of those Jacobi forms whose Fourier coefficients $c(n,r,\gamma)$ vanish unless $\ell^{-1} r \in \mathbb{Z}^{N} - B^T \mathbf{S} \gamma.$ \par \

There is a corresponding operator $U_{\ell}$ that acts on modular forms for the Weil representation attached to a (not necessarily direct) sum of two quadratic forms. Let $\mathbf{S}_1 \in \mathbb{Z}^{e \times e}, \mathbf{S}_2 \in \mathbb{Z}^{N \times N}$ be Gram matrices and let $U \in \mathbb{Z}^{e \times N}$. We define $$U_{\ell} = U_{\ell}^{e,N} : M_k \Big( \begin{pmatrix} \mathbf{S}_1 & U \\ U^T & \mathbf{S}_2 \end{pmatrix} \Big) \longrightarrow M_k \Big( \begin{pmatrix} \mathbf{S}_1 & \ell U \\ \ell U^T & \ell^2 \mathbf{S}_2 \end{pmatrix} \Big)$$ as the pullback $\uparrow_{\mathbb{Z}^{e} \times \mathbb{Z}^{N}}^{\mathbb{Z}^{e} \times \ell \mathbb{Z}^{N}}$ in the sense of lemma 5.5 of \cite{Br}; that is, $$U_{\ell} \Big[ \sum_{n,\gamma} c(n,\gamma) q^n \mathfrak{e}_{\gamma} \Big] = \sum_{n,\gamma} c(n,\gamma) q^n \times \sum_{p(\delta) = \gamma} \mathfrak{e}_{\delta},$$ where $p$ is a projection map that amounts to multiplying the second block component by $\ell$. \par \

\begin{ex} The weight three Eisenstein series for the Weil representation attached to the lattice with Gram matrix $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ is $$E_3(\tau) = (1 + 72q + 270q^2 + 720q^3 + ...) \mathfrak{e}_{(0,0)} + (27q^{2/3} + 216q^{5/3} + 459q^{8/3} +...) (\mathfrak{e}_{(1/3,1/3)} + \mathfrak{e}_{(2/3,2/3)}).$$ Letting $e = N = 1$, its image under the Hecke map $U_2$ is \begin{align*} U_2 E_3(\tau) &= (1 + 72q + 270q^2 + ...) (\mathfrak{e}_{(0,0)} + \mathfrak{e}_{(0,1/2)}) + \\ &\quad +  (27q^{2/3} + 216q^{5/3} + 459q^{8/3} + ...) (\mathfrak{e}_{(1/3,2/3)} + \mathfrak{e}_{(2/3,1/3)} + \mathfrak{e}_{(1/3,1/6)} + \mathfrak{e}_{(2/3,5/6)}), \end{align*} which is a weight three modular form for the Weil representation attached to the Gram matrix $\begin{pmatrix} 2 & 2 \\ 2 & 8 \end{pmatrix}$.
\end{ex}

\begin{rem} One can define $U_{\ell}$ on arbitrary (holomorphic) functions $f(\tau,z)$ by $U_{\ell} f(\tau,z)= f(\tau,\ell z)$. Then $U_{\ell}$ respects the Petersson slash operator in the sense that $$U_{\ell} \Big( f|_{k,\mathcal{M},B} M \Big) = (U_{\ell} f) \Big|_{k,\ell^2 \mathcal{M}, \ell B} M$$ and $$U_{\ell} \Big( f |_{k,\mathcal{M},B} (\ell \lambda, \ell \mu, \ell^2 t) \Big) = (U_{\ell} f) \Big|_{k,\ell^2 \mathcal{M}, \ell B} (\lambda, \mu, t).$$ A similar statement holds for the operator $U_{\ell}$ on modular forms.
\end{rem}

\section{Theta decomposition}

Let $\mathbf{S}_1$ and $\mathbf{S}_2$ be Gram matrices of sizes $e,N$ where $\mathbf{S}_2$ is positive-definite and set $\mathcal{M} = \frac{1}{2}\mathbf{S}_2$. There is an isomorphism of modules $$M_{k-N/2}(\mathbf{S}_1 \oplus \mathbf{S}_2) \longrightarrow J_{k,\mathcal{M},0}(\mathbf{S}_1)$$ over the ring $M_*$ of scalar modular forms which is essentially given by tensoring with the theta function $\Theta_{\mathbf{S}_2}(\tau,z)$ of $\mathbf{S}_2$ and applying a trace map (see the next section for more details) and which identifies the modular form $$F(\tau) = \sum_{\gamma \in A(\mathbf{S}_1 \oplus \mathbf{S}_2)} \sum_n c(n,\gamma) q^n \mathfrak{e}_{\gamma}$$ with the Jacobi form $$\Phi(\tau,z) = \Theta(F)(\tau,z) = \sum_{\gamma \in A(\mathbf{S}_1)} \sum_{n,r} c(n,(\gamma,\mathbf{S}_2^{-1} r)) q^{n + r^T \mathbf{S}_2^{-1} r / 2} \zeta^r \mathfrak{e}_{\gamma}.$$

In this section we record a simple generalization of this. Let $\mathbf{S}$ be a Gram matrix of size $e$ and let $B \in \mathbb{Q}^{e \times N}$ be a matrix for which $\mathbf{S}B$ is integral. Also let $\mathcal{M} \in \mathbb{Q}^{N \times N}$ be a symmetric positive-definite matrix for which $2\mathcal{M} + B^T \mathbf{S}B$ is a Gram matrix, i.e. integral with even diagonal, and set $$\mathbf{\tilde S} = \begin{pmatrix} \mathbf{S} & \mathbf{S} B \\ B^T \mathbf{S} & 2 \mathcal{M} + B^T \mathbf{S} B \end{pmatrix}.$$

\begin{prop} There is an isomorphism of $M_*$-modules $$\Theta : M_{*-N/2}(\mathbf{\tilde S}) \longrightarrow J_{*,\mathcal{M},B}(\mathbf{S})$$ that sends a modular form $$F(\tau) = \sum_{\gamma \in A(\mathbf{\tilde S})} \sum_{n \in \mathbb{Z}^{e + N} - \tilde Q(\gamma)} c(n,\gamma) q^n \mathfrak{e}_{\gamma}$$ to the Jacobi form $$\Phi(\tau,z) = \sum_{\gamma \in A(\mathbf{S})} \sum_{n \in \mathbb{Z} - Q(\gamma)} \sum_{r \in \mathbb{Z}^{N} - B^T \mathbf{S}\gamma} c\Big(n, (\gamma - \frac{1}{2} B \mathcal{M}^{-1} r, \frac{1}{2} \mathcal{M}^{-1} r)\Big) q^{n + \frac{1}{4} r^T \mathcal{M}^{-1} r} \zeta^r \mathfrak{e}_{\gamma}.$$
\end{prop}
\begin{proof} First we should clarify that if $\gamma \in \mathbf{S}^{-1} \mathbb{Z}^e$ and $r \in \mathbb{Z}^N - B^T \mathbf{S} \gamma$, then $$\begin{pmatrix} \mathbf{S} & \mathbf{S} B \\ B^T \mathbf{S} & 2 \mathcal{M} + B^T \mathbf{S} B \end{pmatrix} \begin{pmatrix} \gamma - \frac{1}{2} B \mathcal{M}^{-1} r \\ \frac{1}{2} \mathcal{M}^{-1} r \end{pmatrix} = \begin{pmatrix} \mathbf{S} \gamma \\ B^T \mathbf{S} \gamma + r \end{pmatrix} \in \mathbb{Z}^{e+N}$$ and therefore $$\left(\gamma - \frac{1}{2}B \mathcal{M}^{-1} r, \frac{1}{2} \mathcal{M}^{-1} r\right) \in \mathbf{\tilde S}^{-1} \mathbb{Z}^{e+N};$$ moreover this is a well-defined element of $A(\mathbf{\tilde S})$ when $\gamma$ is interpreted as an element of $A(\mathbf{S})$. \par

Choose $\ell \in \mathbb{N}$ such that $\ell \cdot B \in \mathbb{Z}^{e \times N}$. The Gram matrix $\begin{pmatrix} \mathbf{S} & \ell \mathbf{S} B \\ \ell B^T \mathbf{S} & \ell^2 (2 \mathcal{M} + B^T \mathbf{S} B) \end{pmatrix}$ is equivalent to $\mathbf{S} \oplus 2 \ell^2 \mathcal{M}$ so for every weight $k$ we have an injective map $$M_{k-N / 2}(\mathbf{\tilde S}) \stackrel{U_{\ell}}{\longrightarrow} M_{k-N / 2}(\mathbf{S} \oplus 2 \ell^2 \mathcal{M})\longrightarrow J_{k,\ell^2 \mathcal{M},0}(\mathbf{S}).$$ In terms of Fourier expansions this sends $F(\tau) = \sum_{(\gamma_1,\gamma_2) \in A(\mathbf{\tilde S})} \sum_n c(n,(\gamma_1,\gamma_2)) q^n \mathfrak{e}_{(\gamma_1,\gamma_2)}$ first to the modular form $$U_{\ell} F(\tau) = \sum_{\gamma_1 \in A(\mathbf{S})} \sum_{\gamma_2 \in A(2 \ell^2 \mathcal{M})} \sum_n c(n,(\gamma_1 - \ell B\gamma_2, \ell \gamma_2)) q^n \mathfrak{e}_{(\gamma_1, \gamma_2)}$$ and then to the Jacobi form $$\Phi(\tau,z) = \sum_{\gamma \in A(\mathbf{S})} \sum_{n,r}  c\Big(n,\left(\gamma - \frac{1}{2\ell} B \mathcal{M}^{-1} r , \frac{1}{2\ell} \mathcal{M}^{-1} r\right)\Big) q^{n + \frac{1}{4\ell^2} r^T \mathcal{M}^{-1} r} \zeta^r \mathfrak{e}_{\gamma}.$$ These coefficients vanish unless $$\begin{pmatrix} \mathbf{S} & \mathbf{S}B \\ B^T \mathbf{S} & 2 \mathcal{M} + B^T \mathbf{S} B \end{pmatrix} \begin{pmatrix} \gamma - \frac{1}{2\ell} B \mathcal{M}^{-1} r \\ \frac{1}{2\ell} \mathcal{M}^{-1} r \end{pmatrix} = \begin{pmatrix} \mathbf{S} \gamma \\ B^T \mathbf{S} \gamma + \ell^{-1} r \end{pmatrix}$$ is integral, or equivalently $\ell^{-1} r \in \mathbb{Z}^{N} - B^T \mathbf{S} \gamma$, which is the condition for $\Phi$ to be in the range of $U_{\ell} : J_{k,\mathcal{M},B}(\mathbf{S}) \rightarrow J_{k,\ell^2 \mathcal{M},0}(\mathbf{S})$. Therefore we define the theta decomposition by requiring the diagram

\begin{center}
\begin{tikzpicture}[node distance=2cm, auto]
\node (A) {$M_{k-N/2}(\mathbf{\tilde S})$};
\node(AA) [right of = A] {$ $};
\node (B) [right of = AA] {$M_{k-N/2}(\mathbf{S} \oplus 2 \ell^2 \mathcal{M})$};
\node (C) [below of = A] {$J_{k,\mathcal{M},B}(\mathbf{S})$};
\node (D) [below of = B] {$J_{k,\ell^2 \mathcal{M},0}(\mathbf{S})$};
\draw[dashed,->] (A) to node {$ $} (C);
\draw[->] (A) to node {$U_{\ell}$} (B);
\draw[->] (C) to node {$U_{\ell}$} (D);
\draw[->] (B) to node {$ $} (D);
\end{tikzpicture}
\end{center} to commute. Its inverse can be constructed in the same way: starting with a Jacobi form, apply $U_{\ell}$ and the usual theta decomposition, then identify the result in the image of $U_{\ell}$ on modular forms. That this is independent of $\ell$ can be seen from the coefficient formula for $\Phi$ in the claim.
\end{proof}

\section{Eisenstein and Poincar\'e series}

Some of the simplest Jacobi forms that can be constructed are the Poincar\'e series: $$\mathcal{P}_{k,(\mathcal{M},B),(n,r,\gamma)}(\tau) = \sum_{(\lambda,0,0,M) \in \mathcal{J}_{N,\infty} \backslash \mathcal{J}_{N}} q^n \zeta^r \mathfrak{e}_{\gamma} \Big|_{k,\mathcal{M},B} (\zeta,M),$$ where $(\lambda,0,0,M)$ runs through cosets of the Jacobi group $\mathcal{J}_{N}$ by the subgroup generated by all $(0,\mu,t) \in \mathcal{H}_{N}$ and by $T = (\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, 1)$ and $Z = (-I,i)$ from $Mp_2(\mathbb{Z})$. The series $P_{n,r,\gamma}(\tau)$ is well-defined as long as $n \in \mathbb{Z} - Q(\gamma)$ and $r \in \mathbb{Z}^{N} - B^T \mathbf{S}\gamma$ and $k > 2 + N/2$ and it always transforms like a Jacobi form by construction. The vanishing condition on Fourier coefficients is satisfied when $4n - r^T \mathcal{M}^{-1} r \ge 0$. If $n=r=0$ then we call $\mathcal{E}_{k,\mathcal{M},\gamma}(\tau) = \mathcal{P}_{k,\mathcal{M},0,0,\gamma}(\tau)$ the \textbf{Eisenstein series}. See also \cite{M2} which considers the scalar-valued case in more detail. \par \

The point of this section is to show that the Jacobi Eisenstein and Poincar\'e series as above correspond to the usual Eisenstein and Poincar\'e series as in chapter $1$ of \cite{Br}: $$P_{k,n,\gamma}(\tau) = \sum_{M \in \Gamma_{\infty} \backslash \Gamma} q^n \mathfrak{e}_{\gamma} \Big|_{k,\rho^*} M;$$ i.e. any coefficient formula for Jacobi Eisenstein and Poincar\'e series even in the generality considered here (which includes the author's computations in \cite{W}) will reduce to theorem 1.3 of \cite{Br}. \par \

\begin{prop} In the notation of the previous section, the theta decomposition satisfies $$\Theta \Big( P_{k-N / 2, n, (\gamma - \frac{1}{2} B \mathcal{M}^{-1} r, \frac{1}{2} \mathcal{M}^{-1} r)} \Big) = \mathcal{P}_{k,(\mathcal{M},B),(n + r^T \mathcal{M}^{-1} r / 4, r, \gamma)}.$$
\end{prop}
Poincar\'e series extract Fourier coefficients with respect to an appropriately defined Petersson scalar product so this correspondence is suggested by the formula in proposition 7; one could give a proof along these lines but we use a more direct argument here.
\begin{proof} Suppose first that $B=0$ so the theta decomposition maps $F \in M_{k-N/2}(\mathbf{S} \oplus 2\mathcal{M})$ to $\pi(F \otimes \Theta)$ where $\Theta(\tau,z) = \sum_{\lambda \in \mathbb{Z}^{N}} q^{\lambda^T \mathcal{M}^{-1} \lambda / 4} \zeta^{\lambda} \mathfrak{e}_{(2\mathcal{M})^{-1} \lambda}$ and where $\pi$ is the linear map $$\pi : \mathbb{C}[A(\mathbf{S})] \otimes \mathbb{C}[A(2\mathcal{M})] \otimes \mathbb{C}[A(2\mathcal{M})] \rightarrow \mathbb{C}[A(\mathbf{S})], \quad \pi(\mathfrak{e}_{\beta} \otimes \mathfrak{e}_{\gamma} \otimes \mathfrak{e}_{\delta}) = \begin{cases} \mathfrak{e}_{\beta}: & \gamma = \delta; \\ 0: & \text{otherwise}. \end{cases}$$ Here $\Theta$ is a Jacobi form of weight $N / 2$ and index $\mathcal{M}$ for the (non-dual) Weil representation $\rho_{2\mathcal{M}}$ by Poisson summation. One can check that $\pi$ satisfies $\pi \circ (\rho_{\mathbf{S} \oplus 2 \mathcal{M}}^* \otimes \rho_{2\mathcal{M}}(M)) = \rho_{\mathbf{S}}^*(M) \circ \pi$ for all $M \in Mp_2(\mathbb{Z})$ on the generators $M = S,T$. It follows that if $F(\tau) = \sum_{M \in \Gamma_{\infty} \backslash \Gamma} \phi(\tau) \Big|_{k - N / 2, \rho_{\mathbf{S} \otimes 2 \mathcal{M}}^*} M$ for some seed function $\phi$ then

\begin{align*} &\quad \sum_{M \in \Gamma_{\infty} \backslash \Gamma} \pi\Big( \phi(\tau) \otimes \Theta(\tau,z) \Big) \Big|_{k,\mathcal{M}, \rho_{\mathbf{S}}^*} M \\ &= \pi\sum_{M \in \Gamma_{\infty} \backslash \Gamma} \Big(\phi(\tau) \otimes \Theta(\tau,z)\Big) \Big|_{k,\mathcal{M}, \rho_{\mathbf{S} \oplus 2 \mathcal{M}}^* \otimes \rho_{2\mathcal{M}}} M \\ &= \pi \sum_{c,d} (c \tau + d)^{-k+N/2} \rho_{\mathbf{S} \oplus 2 \mathcal{M}}^*(M)^{-1} \phi(M \cdot \tau) \otimes (c \tau + d)^{-N/2} \mathbf{e}\Big(-\frac{c}{c\tau + d} z^T \mathcal{M} z \Big) \rho_{2\mathcal{M}}(M)^{-1} \Theta\left(\frac{a\tau + b}{c \tau + d},\frac{z}{c\tau + d}\right) \\ &= \pi \sum_{c,d} (c \tau + d)^{-k+N/2} \rho_{\mathbf{S} \oplus 2 \mathcal{M}}^*(M)^{-1} \phi(M \cdot \tau) \otimes \Theta(\tau,z) \\ &= \pi \Big(F(\tau) \otimes \Theta(\tau,z)\Big) \end{align*} is the Jacobi form corresponding to $F$. When $\phi(\tau) = q^n \mathfrak{e}_{\gamma} \otimes \mathfrak{e}_{(2 \mathcal{M})^{-1} r}$ for some $\gamma \in A(\mathbf{S})$ then

\begin{align*} \pi\Big( \phi(\tau) \otimes \Theta(\tau,z) \Big) &= \sum_{\mu \in r + 2 \mathcal{M} \mathbb{Z}^{N}} q^{n + \mu^T \mathcal{M}^{-1} \mu / 4} \zeta^{\mu} \mathfrak{e}_{\gamma} \\ &= \sum_{\lambda \in \mathbb{Z}^{N}} \mathbf{e}\Big( \tau \lambda^T \mathcal{M} \lambda + 2\lambda^T \mathcal{M} z \Big) q^{n + r^T \mathcal{M}^{-1} r / 4 + r^T \lambda} \zeta^r \mathfrak{e}_{\gamma} \quad (\mu = r + 2 \mathcal{M} \lambda) \\ &= \sum_{\lambda \in \mathbb{Z}^{N}} q^{n + r^T \mathcal{M}^{-1} r / 4} \zeta^r \mathfrak{e}_{\gamma} \Big|_{k,\mathcal{M}} (\lambda,0,0)\end{align*} and therefore \begin{align*} \pi \Big( P_{k-N/2,n,(\gamma,(2 \mathcal{M})^{-1} r)}(\tau) \otimes \Theta(\tau,z) \Big) &= \sum_{M \in \Gamma_{\infty} \backslash \Gamma} \pi\Big( \phi(\tau) \otimes \Theta(\tau,z) \Big) \Big|_{k,\mathcal{M}, \rho_{\mathbf{S}}^*} M \\ &= \sum_{(\lambda,0,0,M) \in \mathcal{J}_{\infty} \backslash \mathcal{J}} \Big( q^{n + r^T \mathcal{M}^{-1} r / 4} \zeta^r \mathfrak{e}_{\gamma} \Big) \Big|_{k,\mathcal{M}} (\lambda,0,0) \Big|_{k,\mathcal{M},\rho_{\mathbf{S}}^*} M \\ &= \mathcal{P}_{k,\mathcal{M},(n+r^T \mathcal{M}^{-1} r / 4, r, \gamma)}(\tau,z). \end{align*}

For the general case of theta decomposition in the Gram matrix $\mathbf{\tilde S} = \begin{pmatrix} \mathbf{S} & \mathbf{S} B \\ B^T \mathbf{S} & 2 \mathcal{M} + B^T \mathbf{S} B \end{pmatrix}$ we need to check that the images of $P_{k-N/2,n,(\gamma-\frac{1}{2}B \mathcal{M}^{-1} r, \frac{1}{2} \mathcal{M}^{-1} r)}$ and $\mathcal{P}_{k,(\mathcal{M},B),(n + r^T \mathcal{M}^{-1} r / 4, r, \gamma)}$ match up correctly under $U_{\ell}$. On Jacobi forms we find \begin{align*} &\quad U_{\ell} \mathcal{P}_{k,(\mathcal{M},B),(n + r^T \mathcal{M}^{-1} r/4,r,\gamma)} \\ &= U_{\ell} \sum_{M \in \Gamma_{\infty} \backslash \Gamma} \sum_{\lambda \in \mathbb{Z}^N} (q^{n + r^T \mathcal{M}^{-1} r / 4} \zeta^r \mathfrak{e}_{\gamma}) \Big|_{k,\mathcal{M},B} (\lambda,0,0) \Big|_{k,\mathcal{M},B,\rho_{\mathbf{S}}^*} M \\ &= \sum_{M \in \Gamma_{\infty} \backslash \Gamma} \sum_{\lambda \in \mathbb{Z}^N} \Big[ U_{\ell} \Big( \sum_{\mu \in \mathbb{Z}^N / \ell \mathbb{Z}^N} q^{n + r^T \mathcal{M}^{-1} r / 4} \zeta^r \mathfrak{e}_{\gamma} \Big|_{k,\mathcal{M},B} (\mu,0,0) \Big) \Big] \Big|_{k,\ell^2\mathcal{M},\ell B} (\lambda,0,0) \Big|_{k,\ell^2 \mathcal{M},\ell B,\rho_{\mathbf{S}}^*} M \\ &= \sum_{\mu \in \mathbb{Z}^N / \ell \mathbb{Z}^N} \mathcal{P}_{k, \ell^2 \mathcal{M}, (n + (r + 2 \mathcal{M}\mu)^T \mathcal{M}^{-1}(r + 2 \mathcal{M}\mu)/4, \ell r + 2\ell \mathcal{M}\mu, \gamma + B \mu)}, \end{align*} while on modular forms, abbreviating $\tilde \gamma = (\gamma - \frac{1}{2}B \mathcal{M}^{-1} r, \frac{1}{2} \mathcal{M}^{-1} r)$ and applying $\uparrow$ termwise gives \begin{align*} U_{\ell} P_{k-N/2,n,\tilde \gamma} &= U_{\ell} \Big[ \sum_{M \in \Gamma_{\infty} \backslash \Gamma} q^n \mathfrak{e}_{\tilde \gamma} \Big|_{k,\rho_{\mathbf{\tilde S}}^*} M \Big] \\ &= \sum_{M \in \Gamma_{\infty} \backslash \Gamma} \Big( q^n \sum_{\substack{\delta \in A(\mathbf{S} \oplus 2 \ell^2 \mathcal{M}) \\ p(\delta) = \tilde \gamma}} \mathfrak{e}_{\delta} \Big) \Big|_{k,\rho_{\mathbf{S} \oplus 2 \ell^2 \mathcal{M}}^*} M \\ &= \sum_{\mu \in \mathbb{Z}^N / \ell \mathbb{Z}^N} P_{k-N/2,n,(\gamma - \frac{1}{2} \ell^{-1} \mathcal{M}^{-1} r, \frac{1}{2}\ell^{-1} \mathcal{M}^{-1} r + \ell^{-1} \mu)}, \end{align*} which correspond as they are expected to.
\end{proof}

\section{Kohnen plus space}

Since Jacobi forms of very small index tend to be determined completely by their value at $z=0$, applying the theta decomposition and setting $z=0$ in the resulting Jacobi forms gives isomorphisms between modular forms of apparently unrelated levels. The simplest example is an identification of the Kohnen plus spaces of half-integral weight modular forms of level $4$ with modular forms of level $3$ and odd integral weight. \par \

Let $k \in \mathbb{N}_0$. The Kohnen plus space of weight $k + 1/2$ is the subspace $M_{k+1/2}^+(\Gamma_0(4))$ of modular forms of level $\Gamma_0(4)$ in which the Fourier coefficient $a_n$ is zero unless $(-1)^k n \equiv 0,1$ mod $4$. This was introduced by Kohnen in \cite{K} (also for more general levels) as a natural setting for the Shimura correspondence: in particular there are Hecke-equivariant isomorphisms $M_{k+1/2}^+ \cong M_{2k}$ to classical modular forms of integral weight and level $1$. \par \

\begin{prop} For every odd integer $k$, there is an isomorphism $$M_k(\Gamma_1(3)) = M_k(\Gamma_0(3),\chi) \cong M_{k-1/2}^+(\Gamma_0(4)) \times M_{k+1/2}^+(\Gamma_0(4)) .$$ Here $\chi$ is the nontrivial Nebentypus modulo $3$. More precisely, there are isomorphisms $$M_k^+(\Gamma_0(3),\chi) \cong M_{k-1/2}^+(\Gamma_0(4)), \quad M_k^-(\Gamma_0(3),\chi) \cong M_{k+1/2}^+(\Gamma_0(4)),$$ where $M_k^+(\Gamma_0(3),\chi)$ is the subspace of forms whose coefficients $a_n$ vanish when $n \equiv 2$ mod $3$; and $M_k^-(\Gamma_0(3),\chi)$ is the subspace of forms whose coefficients vanish when $n \equiv 1$ mod $3$.
\end{prop}
These isomorphisms are not Hecke-equivariant but they are simple to express in terms of the coefficients, and more generally they are isomorphisms of graded modules over the ring $M_*$ of classical modular forms of level one; see the remark below.
\begin{proof} Using a more general result of Bruinier and Bundschuh \cite{BB}, all of the spaces here can be interpreted as spaces of vector-valued modular forms: $$M_k^+(\Gamma_0(3),\chi) \cong M_k \Big( \begin{pmatrix} -2 & -1 \\ -1 & -2 \end{pmatrix} \Big), \quad M_k^-(\Gamma_0(3),\chi) \cong M_k\Big( \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \Big),$$ $$M_{k-1/2}^+(\Gamma_0(4)) \cong M_{k-1/2}\Big( \begin{pmatrix} -2 & -1 &1 \\ -1 & -2 & 0 \\ 1 & 0 & 0 \end{pmatrix} \Big), \quad M_{k+1/2}^+(\Gamma_0(4)) \cong M_{k+1/2}((2)),$$ since all of the Gram matrices above have prime determinant. These identifications are essentially given by summing all components; such that for example the weight three Eisenstein series $$E_3(\tau) = (1 - 90q - 216q^2 - ...) \mathfrak{e}_{(0,0)} + (-9q^{1/3} - 117q^{4/3} - ...) (\mathfrak{e}_{(1/3,1/3)} + \mathfrak{e}_{(2/3,2/3)}) \in M_3\Big( \begin{pmatrix} -2 & -1 \\ -1 & -2 \end{pmatrix} \Big)$$ and $$E_3(\tau) = (1 + 72q + 270q^2 + ...) \mathfrak{e}_{(0,0)} + (27q^{2/3} + 216q^{5/3} + ...) (\mathfrak{e}_{(1/3,1/3)} + \mathfrak{e}_{(2/3,2/3)}) \in M_3 \Big( \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \Big)$$ correspond to the scalar modular forms $$1 - 18q - 90q^3 - 234q^4 - 216q^6 - ... \in M_3(\Gamma_0(3),\chi)^+$$ and $$1 + 54q^2 + 72q^3 + 432q^5 + 270q^6 + ... \in M_3(\Gamma_0(3),\chi)^-,$$ respectively. The theta decomposition gives identifications $$M_k\Big( \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \Big) \cong J_{k+1/2,3/4,(1/2)}((2))$$ and $$M_{k-1/2}\Big( \begin{pmatrix} -2 & -1 & 1 \\ -1 & -2 & 0 \\ 1 & 0 & 0 \end{pmatrix}\Big) \cong J_{k,1/3,(-2/3,1/3)} \Big( \begin{pmatrix} -2 & -1 \\ -1 & -2 \end{pmatrix} \Big)$$ and setting $z=0$ in the resulting Jacobi forms gives the maps $$M_k\Big( \begin{pmatrix}2 & 1 \\ 1 & 2 \end{pmatrix} \Big) \rightarrow M_{k+1/2}((2)), \; M_{k-1/2}\Big( \begin{pmatrix} -2 & -1 & 1 \\ -1 & -2 & 0 \\ 1 & 0 & 0 \end{pmatrix} \Big) \rightarrow M_k\Big( \begin{pmatrix} -2 & -1 \\ -1 & -2 \end{pmatrix} \Big)$$ in the claim. Both maps are injective: one can use theorem 1.2 of \cite{EZ} to see that any Jacobi form $\Phi$ of index strictly less than one half for which $\Phi(\tau,0)$ vanishes, or of index less than one for which $\Phi(\tau,0)$ and $\frac{d}{dz}\Big|_{z=0} \Phi(\tau,z)$ both vanish, must be identically zero. (In the second case, $\frac{d}{dz}\Big|_{z=0} \Phi(\tau,z)$ is a modular form of antisymmetric weight for the Gram matrix $(2)$ so it vanishes identically regardless of $\Phi$.) Theorem 1.2 of \cite{EZ} considers only integer-index Jacobi forms but we can reduce to that case by tensoring $\Phi$ with itself enough times to produce something of integer index. \par \

To see that these are isomorphisms we compare the dimensions of both sides. One way to show that the dimensions are the same is to check the first few weights and then use the fact that for $k \ge 3$, the dimensions in all four cases increase by $2$ when we replace $k$ by $k + 12$ (since $2$ is the number of elements $\gamma$ in each discriminant group up to equivalence $\gamma \sim -\gamma$).
\end{proof}

\begin{rem}
Unraveling this argument in terms of scalar-valued modular forms, the isomorphisms above are $$M_k^-(\Gamma_1(3)) \rightarrow M_{k+1/2}^+(\Gamma_0(4)), \quad f(\tau) \mapsto \frac{1}{18} \begin{pmatrix} f(\frac{4\tau}{3}) & f(\frac{4 \tau + 1}{3}) & f(\frac{4 \tau + 2}{3}) \end{pmatrix} \begin{pmatrix} 4 & 1 & 1 \\ 1 & 4 & 1 \\ 1 & 1 & 4 \end{pmatrix} \begin{pmatrix} \vartheta(\frac{\tau}{3}) \\ \vartheta(\frac{\tau + 1}{3}) \\ \vartheta(\frac{\tau + 2}{3}) \end{pmatrix}$$ and $$M_{k-1/2}^+(\Gamma_0(4)) \rightarrow M_k^+(\Gamma_1(3)),$$ $$f(\tau) \mapsto \frac{1}{4} \sum_{a \in \mathbb{Z}/4\mathbb{Z}} f\Big( \frac{3 \tau + a}{4} \Big) \vartheta\Big( \frac{\tau - a}{4} \Big),$$ where $\vartheta(\tau) = 1 + 2q + 2q^4 + 2q^9 + ...$ One could probably prove directly that these are well-defined isomorphisms but using the argument above this is more or less automatic.
\end{rem}

\begin{ex} The first part of remark 10 means that to compute the Kohnen plus form corresponding to $f(\tau) = \sum_n a_n q^n \in M_k^-(\Gamma_1(3))$, we divide all coefficients $a_n$, $n \not \equiv 0 \, (3)$ by two, change variables $q \mapsto q^{4/3}$, multiply by $\vartheta(\tau/3)$, and restrict to integer exponents. For example, to compute the Kohnen plus form corresponding to $1 + 54q^2 + 72q^3 + 432q^5 + 270q^6 + ...$ we multiply \begin{align*} &\quad \Big( 1 + 27q^{8/3} + 72q^4 + 216q^{20/3} + 270q^8 + ... \Big) \Big( 1 + 2q^{1/3} + 2q^{4/3} + 2q^3 + ... \Big) \\ &= 1 + 2q^{1/3} + 2q^{4/3} + 27q^{8/3} + 56q^3 +  126q^4 + 144q^{13/3} + 146q^{16/3} + 54q^{17/3} + 216q^{20/3} + 576q^7 + ... \end{align*} and restrict to integer exponents to find the Cohen Eisenstein series $$1 + 56q^3 + 126q^4 + 576q^7 + ...$$ of weight $7/2$ \cite{C}. \par \

The second part of remark 10 means that to compute the level three form corresponding to a Kohnen plus form $f(\tau)$, we multiply $f(3\tau/4)$ by $\vartheta(\tau/4)$ and restrict to integer exponents. For example, the Cohen Eisenstein series of weight $5/2$ is $$f(\tau) = 1 - 10q - 70q^4 - 48q^5 - 120q^8 - 250q^9 - 240q^{12} - 240q^{13} - ...$$ and since \begin{align*} &\quad \Big( 1 - 10q^{3/4} - 70q^3 - 48q^{15/4} - 120q^6 - 250q^{27/4} - 240q^9 - ... \Big) \Big( 1 + 2q^{1/4} + 2q + 2q^{9/4} + 2q^4 + ... \Big) \\ &= 1 + 2q^{1/4} -10q^{3/4} - 18q - 20q^{7/4} + 2q^{9/4} - 90q^3 - 140q^{13/4} - 48q^{15/4} - 234q^4 - ...\end{align*} we see that the corresponding level three form is $$1 - 18q - 90q^3 - 234q^4 - ... \in M_3^+(\Gamma_1(3)).$$ We could also point out that the theta function $\vartheta(\tau) = 1 + 2q + 2q^4 + 2q^9 + ... \in M_{1/2}^+(\Gamma_0(4))$ itself corresponds to the theta function $\theta(\tau) = 1 + 6q + 6q^3 + 6q^4 + 12q^7 + ... \in M_1^+(\Gamma_0(3),\chi)$ of $x^2 + xy + y^2$ under this isomorphism.

\end{ex}

\section{Hermitian modular products of small weight}

Recall that the Borcherds lift takes vector-valued modular forms for the (non-dual) Weil representation attached to a lattice of signature $(b^+,2)$ with singularities at cusps to orthogonal modular forms with product expansions and known divisors (\cite{B},\cite{Br}). Given such an automorphic product one can produce automorphic forms on subgrassmannians essentially by restriction, with a renormalization process to deal with a possible zero or pole there. This operation is called the quasi-pullback (see \cite{M}, section 3) and it can be used to construct infinite families of Borcherds products. For example, Gritsenko and Nikulin constructed a paramodular form of weight $12$ in every level $t$ except $t=1$ using this kind of argument for the lift of $\Delta^{-1}$ to the Grassmannian of $II_{26,2}$ (\cite{GN2}, remark 4.4). It is also possible to find automorphic forms on subgrassmannians by tensoring the original input form by various theta functions and taking the Borcherds lift of this. Ma showed in \cite{M} that these processes lead to the same result, at least when Koecher's principle applies. \par \

Theta decompositions allow one to identify the nearly-holomorphic input functions with weak Jacobi forms, and in the case of integral index, setting $z=0$ in those Jacobi forms has the same effect as tensoring with theta functions as in the last paragraph. Taking theta decompositions into Jacobi forms of nonintegral index and setting $z=0$ gives a similar interpretation of the $\Theta$-contraction of \cite{M}. \par \

We give an example here by showing that for even discriminants there always exist Hermitian modular products of small weight. These are essentially the same as orthogonal modular forms on Grassmannians of signature $(4,2)$ lattices of the form $\Lambda_K = \mathcal{O}_K \oplus II_{2,2}$, where $\mathcal{O}_K$ denotes the ring of integers in an imaginary-quadratic number field $K$ together with its norm-form $N_{K/\mathbb{Q}}$ (see \cite{Dern} for details). This is based on an observation of A. Krieg who suggested this application to me (see \cite{Kr}). \par

\begin{prop} Let $K$ be an imaginary-quadratic field of even discriminant $d_K$ and set $m = -\frac{d_K}{4}$. \\ (i) If $m$ is a sum of three nonzero squares then there is a holomorphic product for $\Lambda_K$ of weight two. \\ (ii) If $m$ is a sum of two nonzero squares then there is a holomorphic product for $\Lambda_K$ of weight three. \\ (iii) If $m=1$ then there is a holomorphic product for $\Lambda_K$ of weight four.
\end{prop}
The cases $m=1,2$ are well-known and were important in finding the structure of the rings of Hermitian modular forms for $K = \mathbb{Q}(i)$ and $K = \mathbb{Q}(\sqrt{-2})$ in \cite{DK1},\cite{DK2}. One can also find the cases $m=5,6$ in the tables of appendix B of \cite{W1}. Those tables also show that no such statement is possible for odd discriminants; for example, the holomorphic products for $\mathbb{Q}(\sqrt{-3})$ and $\mathbb{Q}(\sqrt{-7})$ of smallest weight have weight $9$ and $7$, respectively. Note that (i) and (ii) can both occur for a single $m$, e.g. $m = 17 = 1^2 + 4^2 = 2^2 + 2^2 + 3^2$, where we get holomorphic products of both weights.
\begin{proof} The products in the claim arise as quasi-pullbacks of a singular-weight product on a Grassmannian of type $O(6,2)$; we will show how these arise as Borcherds lifts using the theta decomposition. Suppose first that $m$ has a representation of the form $m = 1 + a^2 + b^2$ with $a,b \in \mathbb{N}_0$. Then the matrix $$\mathbf{S} = \begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 2m & 2a & 2b \\ 0 & 2a & 2 & 0 \\ 0 & 2b & 0 & 2 \end{pmatrix}$$ is the Gram matrix of $x_1^2 + x_2^2 + x_3^2 + x_4^2$ with respect to the basis $(1,0,0,0),(0,1,a,b),(0,0,1,0),(0,0,0,1)$ of $\mathbb{Z}^4$. There is a nearly-holomorphic modular form $F$ of weight $-2$ with principal part $$F(\tau) = 4\mathfrak{e}_{(0,0,0,0)} + q^{-1/4} (\mathfrak{e}_{(\frac{1}{2},0,0,0)} + \mathfrak{e}_{(0,\frac{1}{2},0,0)} + \mathfrak{e}_{(0,0,\frac{1}{2},0)} + \mathfrak{e}_{(0,0,0,\frac{1}{2})}) + ...$$ (Under the Borcherds lift, this produces the singular weight product on the lattice $4A_1 \oplus II_{2,2}$. Up to an involution, i.e. taking the dual lattice and rescaling, this is the case $m=4$ of theorem 8.1 of \cite{Sch}.) Through the theta decomposition we can interpret $F$ as a weak Jacobi form $\Phi(\tau,z)$ of index $\mathcal{M} = \begin{pmatrix} \frac{1+b^2}{m} & -\frac{ab}{m} \\ -\frac{ab}{m} & \frac{1 + a^2}{m} \end{pmatrix}, \, B = \begin{pmatrix} 0 & 0 \\ a/m & b/m \end{pmatrix}$ for the Weil representation attached to the Gram matrix $\begin{pmatrix} 2 & 0 \\ 0 & 2m \end{pmatrix}$, in which we then set $z=0$. By proposition 7 the coefficient of $q^0 \mathfrak{e}_0$ in $\Phi(\tau,0)$ (i.e. twice the weight of the Borcherds lift) is $$4 + \#\left\{r \in \mathbb{Z}^2: \; \frac{1}{4}r^T \mathcal{M}^{-1} r = \frac{1}{4}\right\} = 4 + \#\{(r_1,r_2) \in \mathbb{Z}^2: \; (1+a^2) r_1^2 + 2ab r_1 r_2 + (1+b^2)r_2^2 = 1\};$$ this is $4$ if $a,b$ are both nonzero, and $6$ if one of $a,b$ is zero, and $8$ if both of $a,b$ are zero. \par \

The same argument works in the general case, but it is somewhat messier to write out: we choose a (primitive) vector $v = v_1 \in \mathbb{Z}^3$ with $v^T v = m$, extend it to a basis $\{v_1,v_2,v_3\}$ of $\mathbb{Z}^3$ and apply theta decomposition to the Gram matrix of $x_1^2 + x_2^2 + x_3^2 + x_4^2$ with respect to the basis (in block form) $\{(1,0),(0,v_1),(0,v_2),(0,v_3)\}$ of $\mathbb{Z}^4$.
\end{proof}

\begin{ex} Using proposition 7, the principal parts of these input functions can be read off the principal part of $$F(\tau) = 4\mathfrak{e}_{(0,0,0,0)} + q^{-1/4} (\mathfrak{e}_{(\frac{1}{2},0,0,0)} + \mathfrak{e}_{(0,\frac{1}{2},0,0)} + \mathfrak{e}_{(0,0,\frac{1}{2},0)} + \mathfrak{e}_{(0,0,0,\frac{1}{2})}) + ...$$  We list the first few examples below: \par \

\begin{longtable}{|l|l|l|}
\caption{Principal parts for Hermitian modular products of small weight} \\
\hline
         $m$            & Weight &              Principal part                                                          \\ \hline
$1$ & $4$ & $8 \mathfrak{e}_{(0,0)} + q^{-1/4} (\mathfrak{e}_{(1/2,0)} + \mathfrak{e}_{(0,1/2)})$ \\ \hline
$2$ & $3$ & $6 \mathfrak{e}_{(0,0)} + 2q^{-1/8} (\mathfrak{e}_{(0,1/4)} + \mathfrak{e}_{(0,3/4)}) + q^{-1/4} \mathfrak{e}_{(1/2,0)}$ \\ \hline
$5$ & $3$ & $\begin{aligned}&6 \mathfrak{e}_{(0,0)} + q^{-1/20} (\mathfrak{e}_{(0,1/10)} + \mathfrak{e}_{(0,9/10)} + \mathfrak{e}_{(1/2,2/5)} + \mathfrak{e}_{(1/2,3/5)}) + \\ &\quad + q^{-1/5} (\mathfrak{e}_{(0,1/5)} + \mathfrak{e}_{(0,4/5)}) + q^{-1/4} (\mathfrak{e}_{(1/2,0)} + \mathfrak{e}_{(0,1/2)}) \end{aligned}$ \\ \hline
$6$ & $2$ & $\begin{aligned} &4 \mathfrak{e}_{(0,0)} + 2q^{-1/24} (\mathfrak{e}_{(0,1/12)} + \mathfrak{e}_{(0,5/12)} + \mathfrak{e}_{(0,7/12)} + \mathfrak{e}_{(0,11/12)}) + \\ &\quad + q^{-1/6} (\mathfrak{e}_{(0,1/6)} + \mathfrak{e}_{(0,5/6)}) + q^{-1/4} \mathfrak{e}_{(1/2,0)} \end{aligned}$ \\ \hline
$10$ & $3$ & $\begin{aligned} &6 \mathfrak{e}_{(0,0)} + q^{-1/40} (\mathfrak{e}_{(0,1/20)} + \mathfrak{e}_{(0,9/20)} + \mathfrak{e}_{(0,11/20)} + \mathfrak{e}_{(0,19/20)}) + \\ &\quad + q^{-3/20} (\mathfrak{e}_{(1/2,3/10)} + \mathfrak{e}_{(1/2,7/10)}) + \\ &\quad + q^{-9/40} (\mathfrak{e}_{(0,3/20)} + \mathfrak{e}_{(0,7/20)} + \mathfrak{e}_{(0,13/20)} + \mathfrak{e}_{(0,17/20)}) + \\ &\quad + q^{-1/4} \mathfrak{e}_{(1/2,0)} \end{aligned}$ \\ \hline
$13$ & $3$ & $\begin{aligned} &6 \mathfrak{e}_{(0,0)} + q^{-1/13} (\mathfrak{e}_{(0,1/13)} + \mathfrak{e}_{(0,12/13)}) + \\ &\quad + q^{-9/52} (\mathfrak{e}_{(0,3/26)} + \mathfrak{e}_{(0,23/26)} + \mathfrak{e}_{(1/2,5/13)} + \mathfrak{e}_{(1/2,8/13)}) + \\ &\quad + q^{-3/13} (\mathfrak{e}_{(0,4/13)} + \mathfrak{e}_{(0,9/13)}) + \\ &\quad + q^{-1/4} (\mathfrak{e}_{(0,1/2)} + \mathfrak{e}_{(1/2,0)}) \end{aligned}$ \\ \hline
$14$ & $2$ & $\begin{aligned} &4 \mathfrak{e}_{(0,0)} + q^{-1/56} (\mathfrak{e}_{(0,1/28)} + \mathfrak{e}_{(0,13/28)} + \mathfrak{e}_{(0,15/28)} + \mathfrak{e}_{(0,27/28)}) + \\ &\quad + q^{-1/28} (\mathfrak{e}_{(1/2,5/14)} + \mathfrak{e}_{(1/2,9/14)}) + \\ &\quad + q^{-1/14}(\mathfrak{e}_{(0,1/14)} + \mathfrak{e}_{(0,13/14)}) + \\ &\quad + q^{-1/7} (\mathfrak{e}_{(0,2/7)} + \mathfrak{e}_{(0,5/7)}) + \\ &\quad + q^{-9/56} (\mathfrak{e}_{(0,3/28)} + \mathfrak{e}_{(0,11/28)} + \mathfrak{e}_{(0,17/28)} + \mathfrak{e}_{(0,25/28)}) + \\ &\quad + q^{-1/4} \mathfrak{e}_{(1/2,0)} \end{aligned}$ \\ \hline
$17$ & $2$ & $\begin{aligned} &4 \mathfrak{e}_{(0,0)} + 2q^{-1/17} (\mathfrak{e}_{(0,1/17)} + \mathfrak{e}_{(0,6/17)} + \mathfrak{e}_{(0,11/17)} + \mathfrak{e}_{(0,16/17)}) + \\ &\quad + q^{-9/68} (\mathfrak{e}_{(0,3/34)} + \mathfrak{e}_{(0,31/34)} + \mathfrak{e}_{(1/2,7/17)} + \mathfrak{e}_{(1/2,10/17)}) + \\ &\quad + q^{-1/4} (\mathfrak{e}_{(1/2,0)} + \mathfrak{e}_{(0,1/2)}) \end{aligned}$ \\ \hline
$17$ & $3$ & $\begin{aligned} &6 \mathfrak{e}_{(0,0)} + q^{-1/68} (\mathfrak{e}_{(0,1/34)} + \mathfrak{e}_{(0,33/34)} + \mathfrak{e}_{(1/2,8/17)} + \mathfrak{e}_{(1/2,9/17)}) + \\ &\quad + q^{-2/17} (\mathfrak{e}_{(0,6/17)} + \mathfrak{e}_{(0,11/17)}) + \\ &\quad + q^{-4/17} (\mathfrak{e}_{(0,2/17)} + \mathfrak{e}_{(0,15/17)}) + \\ &\quad + q^{-13/68} (\mathfrak{e}_{(0,9/34)} + \mathfrak{e}_{(0,25/34)} + \mathfrak{e}_{(1/2,4/17)} + \mathfrak{e}_{(1/2,13/17)}) \\ &\quad + q^{-1/4} (\mathfrak{e}_{(1/2,0)} + \mathfrak{e}_{(0,1/2)}) \end{aligned}$ \\ \hline
\end{longtable}

\end{ex}

\textbf{Acknowledgments:} I thank Shaul Zemel for pointing out some errors in the published version of this note. The author is supported by the LOEWE research unit Uniformized Structures in Arithmetic and Geometry.

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\bibliography{\jobname}

\end{document}

config: arxiv


TITLE: Sum of infinite series by considering Maclaurin Series for $e^x$
QUESTION [2 upvotes]: I want to find:
$ \sum_{n = 0}^{ \infty } \frac{1505n + 1506}{3^n(n+1)!}$
We have:
$e^\frac{x}{3} = \sum_{n = 0}^{ \infty } \frac{x^n}{3^n(n!)}$
which we could integrate on both sides to get:
$ \int e^\frac{x}{3} dx =  \sum_{n = 0}^{ \infty } \frac{ x^{n+1} }{3^n({n+1}!)}$
Now, I'm just missing the numerator $1505n+1506$, which I cannot for the life of me see how to proceed... Any hints would be appreciated!
$\textbf{Problem resolved using hint provided below:}$
$\sum_{n = 0}^{ \infty } \frac{1505n + 1506}{3^n(n+1)!} = \sum_{n = 0}^{ \infty } \frac{1505(n+1) + 1}{3^n(n+1)!} = \sum_{n = 0}^{ \infty } \frac{1505}{3^nn!} + \sum_{n = 0}^{ \infty } \frac{1}{3^n(n+1)!}$
The first term can be computed using:
$e^\frac{x}{3} = \sum_{n = 0}^{ \infty } \frac{x^n}{3^nn!}$
, where we set $x = 1$ to get: $1505e^\frac{1}{3}$
The second term can be computed by taking the definte integral of $e^\frac{x}{3}$:
$ \intop\nolimits_{0}^{1} e^\frac{x}{3} dx =  \sum_{n = 0}^{ \infty } \frac{ 1^{n+1} }{3^n({n+1}!)} - \frac{ 0^{n+1} }{3^n({n+1}!)} = \sum_{n = 0}^{ \infty } \frac{1}{3^n{(n+1)}!} = 3e^\frac{1}{3}-3$
Sum = $1508e^\frac{1}{3}-3$

REPLY [0 votes]: $$S=\sum_{n=0}^{\infty} \frac{an+b}{3^n (n+1)!}= \sum_{n=0}^{\infty} \frac{(a(n+1)+b-a}{3^n (n+1)!}=a\sum_{n=0}^{n}\frac{3^{-n }}{n!}+(b-a)\sum_{n=0}^{\infty} \frac{3^{-n}}{(n+1)!}$$
$$\implies S=ae^{-3}+3(b-a)\sum_{n=0}^{\infty}\frac{3^{-(n+1)}}{(n+1)!}=ae^{-3}+3(b-a)[e^{-3}-1].$$


TITLE: $AB$ is not invertible
QUESTION [5 upvotes]: Is it true that if $A$ is an $m \times n$ matrix and $B$ is an $n \times m$ matrix, with $m > n $ then $\det(AB)=0$?

REPLY [0 votes]: Basically the same as cooper.hat's answer with lower technology.
Consider the system $$B\mathbf x = 0.$$ Since $B$ is an $n\times m$ matrix and $n\lt m$, this system has infinite solutions. So there is one non trivial solution say $\mathbf y$. But then $\mathbf y$ is also a nontrivial solution to the system $$(AB)\mathbf x = 0$$
and so $AB$ is not invertible.
So $\det(AB)=0$.


TITLE: Which is the solution of the limit, $\lim_{x \rightarrow \infty} x\sin (\frac{1}{x})$
QUESTION [1 upvotes]: Which is the solution of the limit,
$$
\lim_{x \rightarrow \infty} x\sin \left(\frac{1}{x}\right)
$$

REPLY [0 votes]: Since $\sin \frac1x \sim \frac1x$ as $x\to \infty$, then $\lim_{x\to \infty}x\sin \frac1x = \lim_{x\to \infty}x \frac1x=1 $


TITLE: Sigma-finite measure (countable sequence)
QUESTION [0 upvotes]: Let $(X,\mathcal{A},\mu)$ be a measure space, $g:X \to \mathbb{R}$ a $\mu$-integrable map and $A=X$\ $g^{-1}(0)$.
How to show that it exists a countable sequence $(A_i)_{i \in \mathbb{N}}$ of $\mathcal{A}$ such that
$A=\bigcup\limits_{i=1}^{\infty} A_{i}$ and $\mu(A_i)<\infty, \ \forall n \in \mathbb{N}$?
I got:
Since $A\in\mathcal{A}\Rightarrow (X$\ $g^{-1}(0))^c \in \mathcal{A}$
So $X=A \dot\cup (X$\ $g^{-1}(0))^c$
By finite additivity of measures: $\mu(X)=\mu(A)+\mu(X$\ $g^{-1}(0))^c$
By non-negativity: $\mu(A)<\infty$
I'm not sure if this works for $A_i$ and how to show that $(A_i)_{i \in \mathbb{N}}$ exists with $A=\bigcup\limits_{i=1}^{\infty} A_{i}$.

REPLY [1 votes]: You can take $A_i=\{x\in X\mid |g(x)|\geq\frac1i\}$.
Observe that $\mu(A_i)=\int\mathbf1_{A_i}\;d\mu\leq\int i\mathbf|g|\;d\mu\leq i\int\mathbf|g|\;d\mu<\infty$.
Also the $A_i$ are preimages of Borel sets wrt measurable $|g|$ so are elements of $\mathcal A$.
It is evident that $\bigcup_{i=1}^{\infty}A_i=\{x\in X\mid g(x)\neq0\}=X\setminus g^{-1}(\{0\})=A$.

REPLY [0 votes]: The function $g$ here doesn't care so much: the space is $\sigma$-finite, so there is a countable cover of measurable sets $(X_j)$ of $X$ such that $\mu(X_j)<\infty$ for each $j\in\Bbb N$.
Then the sequence defined by $X_j\cap A$ cover $A$ and necessarily $\mu(X_j\cap A)\le\mu(X_j)$ because any measure is increasing, that is, if $A\subset B$ then $\mu(A)\le\mu(B)$.