TITLE: Bijections $\mathsf{Sh}(1, p+q)\times \mathsf{Sh}(p, q)\simeq \mathsf{Sh}(p+1, q)\times \mathsf{Sh}(1, P)$?
QUESTION [1 upvotes]: For integers $p, q\geq 1$ let us write $S_{p+q}$ for the set of bijections of the set $I_{p+q}:=\{1, \ldots, p+q\}$. This is a group with the usual composition of maps. Let us then define $$\mathsf{Sh}(p, q):=\{\sigma\in\mathsf{S}_{p+q}: \sigma(1)<\ldots<\sigma(p)\quad \textrm{and}\quad \sigma(p+1)<\ldots<\sigma(p+q)\}. $$ The elements in $\mathsf{Sh}(p, q)$ are called $(p, q)$-shuffles. It is easily seen that the cardinality of $|\mathsf{Sh}(p, q)|$ of $\mathsf{Sh}(p, q)$ is $$|\mathsf{Sh}(p, q)|=\binom{p+q}{q}. $$
So counting elements we see there are bijections:
$$\mathsf{Sh}(1, p+q)\times \mathsf{Sh}(p, q)\longrightarrow \mathsf{Sh}(p+1, q)\times \mathsf{Sh}(1, p)$$
and
$$\mathsf{Sh}(1, p+q)\times \mathsf{Sh}(p, q)\longrightarrow \mathsf{Sh}(p, q+1)\times \mathsf{Sh}(1, q).$$
Can anyone help me construct those bijections explicitly?
Thanks.

REPLY [1 votes]: One can regard the $(p,q)$-shuffles in this way. Take a line of $p+q$ objects and colour $p$ or them red and $q$ of them blue. The corresponding
permutation lists the indices of the red objects in ascending order followed by the indices of the blue objects.
Take a line of $p+q+1$ objects, colour one of then yellow, $p$
of them brown and $q$ of them pink. If you replace yellow by red
and brown and pink both by blue, then you get a $(1,p+q)$ shuffle.
If you remove the yellow and replace browns and pinks by reds and blues
you get a $(p,q)$ shuffle. We can reverse the procedure. So we get
a bijection between these arrangements and
$\text{Sh}(1,p+q)\times\text{Sh}(p,q)$.
On the other hand we could replace yellow and brown by red and pink by blue, or remove the pinks and replace yellow by red and brown by blue.
This gives a bijection to $\text{Sh}(p+1,q)\times\text{Sh}(1,p)$.
With sufficient labour you could disguise these bijections by expressing
everything in terms of permutations.


TITLE: Complex Conjugate Eigenvectors Proof
QUESTION [3 upvotes]: a.) Prove that if $x ^+_- iy$ is a complex conjugate pair of
  eigenvectors of a real matrix A corresponding to complex conjugate
  eigenvalues $\mu ^+_- iv$ with $v\ne 0$, then x and y are linearly
  independent real vectors.
b.) More generally, if $v_j=x_j {^+_-} iy_j$, $j=1,...,k$ are complex
  conjugate pairs of eigenvectors corresponding to distinct pairs of
  complex conjugate eigenvalues $\mu_j {^+_-} iv_j,v_j \ne 0$, then the
  real vectors $x_1,...,x_k,y_1,...,y_k$ are linearly independent.

How will I be able to prove these?

REPLY [4 votes]: a) Assume that $cx = y$ and $c\in {\bf R}$
$A(x + icx) = (\mu + i \nu)(x+icx)$ so that $Ax =  (\mu + i \nu)x$
Hence $\nu=0$. Contradiction !
b) $Av=cv$ and $Aw = dw$ where $A$ is real, $c$, $d\in {\bf C}$, and $ v\neq 0$, $w\neq 0\in {\bf C}^n$
If $c\neq d$, and if $av + bw=0$, then $0= aAv + b Aw = acv + bdw$ so that
  $-av = bw=bd/cw$ Hence $b=0$ implies $a=0$. $b\neq 0$ implies $d/c=1$ It is a contradiction. So we conculde that eigenvectors corresponded to differen eigenvalues are independent.
Similarly we can show that $n$ eigenvectors corresponded to different $n$ eigenvalues are independent.
Assume that $a_i x_i + b_i y_i =0$. So $(a_k -i b_k) (x_k + iy_k)  + (a_k + ib_k)(x_k-iy_k)  = 0$ This implies that $a_k=b_k=0$ So we finished the proof.


TITLE: Determinants through dot products
QUESTION [6 upvotes]: Thanks for reading!
So, in Paulo Buchsbaum's answer to this question on Quora...
https://www.quora.com/What-is-the-mathematical-intuition-behind-the-determinant-of-a-matrix-How-was-its-definition-conceived-and-why-is-it-important-What-does-it-mean-intuitively
...he defines a beautiful way to calculate the determinant of a matrix through dot products. One I hadn't seen before.
He goes through the process of calculating this determinant for a $(3,3)$  matrix.
I know this is a long request, but I was wondering if someone could take me through the process of doing it for a $(4,4)$ dimensional matrix, as (as you can see from my comment to his answer) I'm getting really confused.
I would really appreciate it!
Thank you!

Edit:
After reading @MatthewTowers comment (thanks), here is my explanation of what he's done for the $(3,3)$ case:
Let's say we've got a $(3,3)$ matrix with a left column vector $\vec{a}$, a middle column vector $\vec{b}$, and a right column vector $\vec{c}$.
These three vectors create a parallelepiped. The volume of that parallelepiped is the determinant of the matrix.
As you read, here's a picture (taken from Paulo's answer) for reference:

Each vector has three components since they exist in three-dimensional space (aka the matrix has three rows).
$\vec{a} = \begin{bmatrix}a_1 \\ a_2 \\ a_3 \end{bmatrix} \vec{b} = \begin{bmatrix}b_1 \\ b_2 \\ b_3 \end{bmatrix} \vec{c} = \begin{bmatrix}c_1 \\ c_2 \\ c_3 \end{bmatrix}$
Note that $\vec{b}$ and $\vec{c}$ create a parallelogram.
This parallelogram is the "base" of the parallelepiped created by all three column-vectors, and the component of $\vec{a}$ perpendicular to this parallelogram is the height of the parallelepiped.
Let's ignore $\vec{a}$ for now.
The first step is to find a vector $\vec{n}$ that's orthogonal to both $\vec{b}$ and $\vec{c}$.
We set $\vec{n} \bullet \vec{b}=0$ and $\vec{n} \bullet \vec{c}=0$.
Or, in other words, $n_1b_1 + n_2b_2 + n_3b_3 = 0$ and $n_1c_1 + n_2c_2 + n_3c_3 = 0$
That's three unknowns and only two equations.
However, we can choose $n_1$ to be whatever we want, which allows us to solve for $\vec{n}$.
The second step is to find a vector $\vec{o}$ that's orthogonal to $\vec{b}$ and $\vec{n}$.
Notice that $\vec{o}$ will be in the same plane as $\vec{b}$ and $\vec{c}$, but will be perpendicular to $\vec{b}$.
We set $\vec{o} \bullet \vec{b}=0$ and $\vec{o} \bullet \vec{n}=0$.
Once more, that gives us three unknowns and only two equations.
However, we can choose $o_1$ to be whatever we want, which allows us to solve for $\vec{o}$.
We now set the magnitude of $\vec{o}$ to be equal to the magnitude of $\vec{b}$.
$\left \| \vec{o} \right \| := \left \| \vec{b} \right \|$
Note that this makes $\vec{o}$ a vector orthogonal to $\vec{b}$ but in the same plane as $\vec{b}$ and $\vec{c}$, whose magnitude is the magnitude of $\vec{b}$.
If we take the dot product of $\vec{o}$ and $\vec{c}$, that gives us the area of the parallelogram created by $\vec{b}$ and $\vec{c}$.
That's because $\vec{c} \bullet \vec{o}$ is the magnitude of the projection of $\vec{c}$ onto $\vec{o}$ multiplied by the magnitude of $\vec{o}$.
The projection of $\vec{c}$ onto $\vec{o}$ is the height of the parallelogram with $\vec{b}$ taken as the base, since $\vec{o}$ is orthogonal to $\vec{b}$.
And, the magnitude of $\vec{o}$ is equal to the magnitude of $\vec{b}$. So, the dot product $\vec{c} \bullet \vec{o}$ is the height times the base, aka the area of the parallelogram created by $\vec{b}$ and $\vec{c}$.
Now, we set the magnitude of $\vec{n}$ equal to the area of that parallelogram.
$\left \| \vec{n} \right \| :=  \vec{c} \bullet \vec{o} $
And finally, we compute $\vec{a} \bullet \vec{n} $, and get the determinant of the matrix!
Why does this last step work?
For the same reason as the previous dot product - $\vec{n}$ is orthogonal to the parallelogram created by $\vec{b}$ and $\vec{c}$.
Taking the dot product of $\vec{a}$ and $\vec{n}$ is the same thing as taking the projection of $\vec{a}$ onto $\vec{n}$ and multiplying it by the magnitude of $\vec{n}$.
The projection of $\vec{a}$ onto $\vec{n}$ is the height of the parallelepiped, with the parallelogram created by $\vec{b}$ and $\vec{c}$ taken to be the base, and the magnitude of $\vec{n}$ is the area of the parallelogram, aka the area of the base.
Once again, height times base equals volume.
The volume of the parallelepiped is the determinant of the matrix!

To conclude, I have two questions:
I can see that what he's doing is getting a matrix with $n$ column vectors, calculating the cross-product of the last $n-1$ vectors, and then taking the dot product of the first vector with that cross product.
However, how does this connect with the "normal" way to calculate determinants? Is there a way to derive the cross-product formula from his method?
Secondly, if someone could just carry out the process for $n=4$, I'd really appreciate it! I'm getting two equations with four unknowns...
Thanks!

Edit 2:
This is just to show what I've done so far, in regards to my second question (the request for someone to take me through this guy's process for a $(4,4)$ matrix):
Calculate $det(\vec{a},\vec{b},\vec{c},\vec{d})$
Take out the $\vec{a}$, and calculate $det(\vec{b},\vec{c},\vec{d})$
To do so, we take out the $\vec{b}$, and calculate $det(\vec{c},\vec{d})$
To do so, we declare an $\vec{n}$ such that $\vec{n} \bullet \vec{c} = 0$ and $\vec{n} \bullet \vec{d} = 0$.
In other words, $\vec{n}$ is orthogonal to the parallelogram created by $\vec{c}$ and $\vec{d}$.
(And here’s where I'm stuck…should $\vec{n}$ be orthogonal to $\vec{a}$ as well? Right now, I have two equations, but four unknowns, and I’m not sure what to do…

REPLY [1 votes]: The method at that site ignores the sign of the determinant, and is phrased in a way that makes the generalization to higher dimensions a bit less clear than I'd like. I'll try here to provide an altered version of the approach up to four dimensions. I'll use bold for vectors, subscripts for components, and double vertical lines for length, so that $\left\Vert \mathbf{a}\right\Vert =\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots}$.
As was mentioned by LutzL in a comment, this method is very closely connected to using the $QR$-decomposition to find the absolute value of the determinant, as described on Wikipedia here.

1D:
Let's calculate $\det\left(\mathbf{a}\right)$ where $\mathbf{a}$ has one nonzero component. It's $\left\Vert \mathbf{a}\right\Vert$  if $\mathbf{a}$ has a positive component and $-\left\Vert \mathbf{a}\right\Vert$  if $\mathbf{a}$ has a negative component.

2D:
Let's calculate $\det\left(\mathbf{a},\mathbf{b}\right)$ where $\mathbf{a}$ and $\mathbf{b}$ are not collinear.
Let's ignore $\mathbf{a}$ for now. The first step is to find a vector $\mathbf{n}$ that's orthogonal to $\mathbf{b}$. We set $\mathbf{n}\bullet\mathbf{b}$ equal to $\boldsymbol{0}$. That's two unknowns and only one equation. In a typical case, the component $n_{1}$ of $\mathbf{n}$ is not forced to be $0$, so it can be whatever we want that's nonzero (e.g. $1$). (In a special case, $n_{1}$ might be forced to be $0$, but then $n_{2}$ can be chosen freely.)
Now scale $\mathbf{n}$ to get a new vector $\mathbf{n}'$ so that $\left\Vert \mathbf{n}'\right\Vert =\left\Vert \mathbf{b}\right\Vert$ . By some geometry, the area of the parallelogram formed by $\mathbf{a}$ and $\mathbf{b}$ is then $\left|\mathbf{n}'\bullet\mathbf{a}\right|$. The determinant is $\pm\mathbf{n}'\bullet\mathbf{a}$ where the $\pm$ sign here (which may not be the sign of the determinant) is positive exactly when the rotation to get from $\mathbf{a}$ to $\mathbf{b}$ is in the same direction (clockwise or counterclockwise) as the rotation to get from $\mathbf{b}$ to $\mathbf{n}'$. Unfortunately, that can't be determined by a dot-product calculation.

3D:
Let's calculate $\det\left(\mathbf{a},\mathbf{b},\mathbf{c}\right)$ where $\mathbf{a},\mathbf{b},\mathbf{c}$ are not coplanar.
Let's ignore $\mathbf{a}$ for now. The first step is to find a vector $\mathbf{n}$ that's orthogonal to both $\mathbf{b}$ and $\mathbf{c}$. We set $\mathbf{n}\bullet\mathbf{b},\mathbf{n}\bullet\mathbf{c}$ equal to $\boldsymbol{0}$. That's three unknowns and only two equations. In a typical case, the component $n_{1}$ of $\mathbf{n}$ is not forced to be $0$, so it can be whatever we want that's nonzero (e.g. $1$). (In a special case, $n_{1}$ might be forced to be $0$, but then $n_{2}$ or $n_{3}$ can be chosen freely.)
The second step is to find a vector $\mathbf{o}$ that's orthogonal to $\mathbf{c}$ (this choice differs from the original author), but lies in the same plane as $\mathbf{b}$ and $\mathbf{c}$. To keep it in that plane, we need $\mathbf{o}$ to be orthogonal to $\mathbf{n}$. So we have $\mathbf{o}\bullet\mathbf{n}=\boldsymbol{0}$ as well as $\mathbf{o}\bullet\mathbf{c}=\boldsymbol{0}$. Again that's three unknowns and two equations, so we have a degree of freedom and could choose a particular value for some component.
Now scale $\mathbf{o}$ to get a new vector $\mathbf{o}'$ so that $\left\Vert \mathbf{o}'\right\Vert =\left\Vert \mathbf{c}\right\Vert$ . By some geometry, the area of the parallelogram formed by $\mathbf{b}$ and $\mathbf{c}$ is then $\left|\mathbf{o}'\bullet\mathbf{b}\right|$. Now scale $\mathbf{n}$ to get a new vector $\mathbf{n}'$ so that $\left\Vert \mathbf{n}'\right\Vert =\left|\mathbf{o}'\bullet\mathbf{b}\right|$. By some geometry, the volume of the parallelepiped formed by $\mathbf{a}, \mathbf{b}, \mathbf{c}$ is then $\left|\mathbf{n}'\bullet\mathbf{a}\right|$. The determinant is then $\pm\mathbf{n}'\bullet\mathbf{a}$ where I'm pretty sure the $\pm$ sign is positive when the handedness (right or left) of $\mathbf{a},\mathbf{b},\mathbf{c}$ is the same as that of $\mathbf{c},\mathbf{n}',\mathbf{o}'$.

4D:
Let's calculate $\det\left(\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}\right)$ where $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}$ are not in the same $3$-dimensional hyperplane.
Let's ignore $\mathbf{a}$ for now. The first step is to find a vector $\mathbf{n}$ that's orthogonal to all three of $\mathbf{b},\mathbf{c},\mathbf{d}$. We set $\mathbf{n}\bullet\mathbf{b},\mathbf{n}\bullet\mathbf{c},\mathbf{n}\bullet\mathbf{c}$ all to $\boldsymbol{0}$. That's four unknowns and only three equations. In a typical case the component $n_{1}$ of $\mathbf{n}$ is not forced to be $0$, so it can be whatever we want that's nonzero (e.g. $1$). (In a special case, $n_{1}$ might be forced to be $0$, but there will be at least one component we could choose freely.)
The second step is to find a vector $\mathbf{o}$ that's orthogonal to $\mathbf{c}$ and $\mathbf{d}$, but lies in the same $3$-dimensional hyperplane as $\mathbf{b},\mathbf{c},\mathbf{d}$. To keep it in that hyperplane, we need $\mathbf{o}$ to be orthogonal to $\mathbf{n}$. So we have $\mathbf{o}\bullet\mathbf{n}=\boldsymbol{0}$ as well as $\mathbf{o}\bullet\mathbf{c},\mathbf{o}\bullet\mathbf{d}=\boldsymbol{0}$. Again that's four unknowns and three equations, so we have a degree of freedom and could choose a particular value for some component.
The third step is to find a vector $\mathbf{p}$ that's orthogonal to $\mathbf{d}$, but lies in the same $2$-dimensional plane as $\mathbf{c},\mathbf{d}$. To keep it in that plane, it should be orthogonal to $\mathbf{n}$ as well as $\mathbf{o}$. So we have $\mathbf{p}\bullet\mathbf{n},\mathbf{p}\bullet\mathbf{o}=\boldsymbol{0}$ as well as $\mathbf{p}\bullet\mathbf{d}=\boldsymbol{0}$. Again that's four unknowns and three equations, so we have a degree of freedom and could choose a particular value for some component.
Now scale $\mathbf{p}$ to get a new vector $\mathbf{p}'$ so that $\left\Vert \mathbf{p}'\right\Vert =\left\Vert \mathbf{d}\right\Vert$ . By some geometry, the area of the parallelogram formed by $\mathbf{c}$ and $\mathbf{d}$ is then $\left|\mathbf{p}'\bullet\mathbf{c}\right|$. Now scale $\mathbf{o}$ to get a new vector $\mathbf{o}'$ so that $\left\Vert \mathbf{o}'\right\Vert =\left|\mathbf{p}'\bullet\mathbf{c}\right|$. By some geometry, the volume of the parallelepiped formed by $\mathbf{b}, \mathbf{c}, \mathbf{d}$ is then $\left|\mathbf{o}'\bullet\mathbf{b}\right|$. Now scale $\mathbf{n}$ to get a new vector $\mathbf{n}'$ so that $\left\Vert \mathbf{n}'\right\Vert =\left|\mathbf{o}'\bullet\mathbf{b}\right|$. By some geometry, the hypervolume of the hyperparallelepiped formed by $\mathbf{a},\mathbf{b}, \mathbf{c}, \mathbf{d}$ is then $\left|\mathbf{n}'\bullet\mathbf{a}\right|$. The determinant is then $\pm\mathbf{n}'\bullet\mathbf{a}$ where I think the $\pm$ sign is positive when the orientation of $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}$ is the same as that of $\mathbf{d},\mathbf{n}',\mathbf{o}',\mathbf{p}'$.


TITLE: A Machinist's Imperfect Disk
QUESTION [3 upvotes]: Exercise

A machinist is required to manufacture a circular metal disk with area $1000$ cm$^2$.

What radius produces such a disk?

If the machinist is allowed an error tolerance of $\pm 5$ cm$^2$ in the area of the disk, how close to the ideal radius in part (1) must the machinist control the radius?

In terms of the $\epsilon$, $\delta$ definition of $\lim \limits_{x \to a}{f(x)} = L$ what is $x$? What is $f(x)$? What value of $\epsilon$ is given? What is the corresponding value of $\delta$?

Solution
Note: I've decided to solve part (3) before part (2), as it helps set the stage of the solution of (2).

1. What radius produces such a disk?

$\pi r_0^2 = 1000 \implies r_0 = \sqrt{\frac{1000}{\pi}} \approx 17.8412$

3. In terms of the $\epsilon$, $\delta$ definition of $\lim \limits_{x \to a}{f(x)} = L$ what is $x$? What is $f(x)$? What value of $\epsilon$ is given? What is the corresponding value of $\delta$?

$a = r_0 \approx 17.8412$
$L = \pi r_0^2 = 1000$
$x = r$
$f(x) = \pi r^2$
$|r - r_0| < \delta$
$|\pi r^2 - \pi r_0^2| < \epsilon$

2. If the machinist is allowed an error tolerance of $\pm 5$ cm$^2$ in the area of the disk, how close to the ideal radius in part (1) must the machinist control the radius?

$|\pi r^2 - \pi r_0^2| < \epsilon$
$\implies |\pi r^2 - 1000| < 5$
$\implies -5 < \pi r^2 - 1000 < 5$
$\implies 995 < \pi r^2 < 1005$
$\implies \frac{995}{\pi} < r^2 < \frac{1005}{\pi}$
$\implies \sqrt{\frac{995}{\pi}} < r < \sqrt{\frac{1005}{\pi}}$
$\implies 17.7966 < r < 17.8858$
$|r - r_0| < \delta \implies |r - 17.8412| < \delta$
$|17.7966 - 17.8412| < \delta \implies 0.0446 < \delta$
$|17.8858 - 17.8412| < \delta \implies 0.0446 < \delta$
$\delta = \min(0.0446, 0.0446) = 0.0446$

Answer

1. What radius produces such a disk?

$$r_0 = 17.8412 \text{ cm}$$

2. If the machinist is allowed an error tolerance of $\pm 5$ cm$^2$ in the area of the disk, how close to the ideal radius in part (1) must the machinist control the radius?

$$\delta = 0.0446 \text{ cm}$$

3. In terms of the $\epsilon$, $\delta$ definition of $\lim \limits_{x \to a}{f(x)} = L$ what is $x$? What is $f(x)$? What value of $\epsilon$ is given? What is the corresponding value of $\delta$?

$$|r - r_0| < \delta$$
$$|\pi r^2 - \pi r_0^2| < \epsilon$$

Request
Is my answer correct? If not, in what part of my solution did I make a mistake?

REPLY [1 votes]: In part 3 you are asked about four items: $x$, $f(x)$, $\epsilon$, and $\delta$. Your answer omits these. To make it clear:

The independent variable $x$ here is the radius $r$ of the disk
The function $f$ maps radius to area, i.e., $r\mapsto \pi r^2$. In other words, $f(x)=\pi x^2$.
The given $\epsilon$ is $5\,\text{cm}^2$
The corresponding (or rather: a suitable) value of $\delta$ is the $\approx 0.0446\,\text{cm}$ you found

On a sidenote, we may observe that picking $\delta$ such that $\frac\epsilon\delta=f'(r_0)$ usually gives a good enough approximation. Indeed, for the problem at hand $f'(x)=2\pi x$, so we quickly obtain $\delta=\frac{\epsilon}{2\pi r_0}\approx 0.0446$.
Things become even easier with relative errors: Raising to $n$th power multiplies (small) relative errors by $n$. Hence for a desired relative output error of $\frac 5{1000}=\frac1{200}$, the relative input error should be bounded by $\frac{1}{400}$, and indeed $\frac1{400}\cdot 17.8412\approx 0.0446$.


TITLE: $J'$ is a fractional ideal, then multiplying by principal ideal does not change class. So one can assume $J^{-1}$ is integral ideal?
QUESTION [0 upvotes]: This is a part of proof given in Pierre Samuel's Algebraic Theory of Numbers which I do not understand why it can be done in this way.
Let $K$ be a number field. Let $A$ be integral closure of $Z$ in $K$. Denote the set of all non-zero fractional ideals as $I(A)$ and non-zero principal fractional ideals as $F(A)$. The ideal class group is denoted as $I(A)/F(A)$.
Let $J'\in I(A)/F(A)$. It is clear that multiplying by a principal ideal of $A$ does not change class. We can assume without loss of generality $J=(J')^{-1}$ is integral ideal($J\subset A$).
Q: Why can I assume $J=(J')^{-1}$ is an integral ideal(i.e $J\subset A$) by multiplying $J'$ by some principal ideal?
If $b\in A, b(J')^{-1}\subset A$, then I consider $(b(J')^{-1})^{-1}=J'(\frac{1}{b})$. There is no guarantee that $\frac{1}{b}\in A$.

REPLY [1 votes]: $\frac{1}{b}$ is not in $A$ unless $b$ is a unit, in which case $(J')^{-1}$ is already integral. But to more directly answer your question, $(\frac{1}{b})$ is not a principal ideal of $A$, but it is a principal fractional ideal of $A$.

REPLY [1 votes]: Let $I' \in I(A)$ be a lift of $J'$, and let $I$ be an integral ideal lift of $(J')^{-1}$, say $I = b(I')^{-1}$. It follows that $II' = bA$ is principal, so $I$ and $I'$ represent inverse elements of the ideal class group.
Note that it does not follow, however, that $I^{-1} \subset A$. The assumption only says $I \subset A$. It is true that, in the ideal class group, $I^{-1}$ could be represented by an integral ideal (e.g., if we assumed WLOG that $I'$ was integral, then $I'$ would be such a representative), but $I^{-1}$ itself is not typically such a representative.
In fact it's not too difficult to show that if $I, I^{-1}$ are both integral ideals, then $I = I^{-1} = A$.


TITLE: How I can prove tht $T$ is isomorphic to a finite set of natural numbers?
QUESTION [0 upvotes]: Let $T$ be a finite abelian group. We can consider $T$ the as group $ℤ/nℤ$ or $ℤ/qℤ×ℤ/mℤ$.
My question is: How I can prove tht $T$ is in bijection with a finite set of natural numbers?
That is, I want to prove that there is a bijection between $T$ and a finite set of natural numbers ${1,2,...,q}$ where $q$ is the finite size of $T$.

REPLY [2 votes]: This is too long to be a comment, so I hope it answers your question:
Lets consider $\mathbb{Z}_n = \mathbb{Z}/n\mathbb{Z}$. We want to construct a bijection with the set $\{1,\ldots,n\}$. Any $a\in\mathbb{Z}_n$ is a non-empty subset of $\mathbb{Z}$ that is closed under addition with multiples of $n$. Since it is closed under addition with arbitrarily large numbers, it must contain some positive number. Define $\phi(a)$ to be the smallest positive number contained in $a$.
Suppose $\phi(a)>n$, then $\phi(a)-n$ is positive and smaller than $\phi(a)$, a contradiction. Thus $\phi\colon \mathbb{Z}_n\to\{1,\ldots,n\}$.
Now it remains to show that $\phi$ is a bijection. Clearly $\phi$ is surjective, since $\phi(\{mn+k|m\in\mathbb{Z}\})=k$ for any $k\in\{1,\ldots,n\}$.
Furthermore $\phi$ in injective, since $\phi(a)=\phi(b)$ implies that $\phi(a)\in a\cap b$ and the elements of $\mathbb{Z}_n$ partition $\mathbb{Z}$.


TITLE: Linearly independence and Smooth Local Section
QUESTION [2 upvotes]: Let $\pi:E\rightarrow M$ a smooth vector bundle, $U$ a neighborhood of $p$ in $M$ and $\sigma_i:U\rightarrow E, i=1,\dots n$ smooth local sections of $E$. If $\{ \sigma_1(p), \dots, \sigma_n(p)\}$ are linearly independent (as elements of the vector space $E_p$), then by continuity they remain linearly independent in some neighborhood of $p$. Can someone please explain to me why this is true?
I assume that the context (vector bundle, sections etc) doesn't play any role, but I am not sure, so I put it in the context I read it.
My attempt was the following:
Define $f_{(a_1,\dots, a_n)}(q)=a_1\sigma_1(q)+\dots+a_n\sigma_n(q)$ on $U$. For a fixed $(a_1,\dots, a_n)\neq 0$, since $f_{(a_1,\dots, a_n)}(p)\neq 0$, we have that there is a neighborhood of $p$ such that $f_{(a_1,\dots, a_n)}(q)\neq 0$ in this neighborhood. But there are infinetely many n-tuples $(a_1,\dots, a_n)\neq 0$ and if I take then intersection of all such neighborhoods there is no guarantee that I will end up with an open neighborhood.
Any ideas?
Thank you in advance

REPLY [4 votes]: Think about $\sigma = \sigma_1 \wedge \dots \wedge \sigma_n$ as a smooth section of $\wedge^n E$. Saying that $\sigma_1(p), \dots, \sigma_n(p)$ are linearly independent is the same as saying that $\sigma(p)$ is non-vanishing. By continuity arguments, the set on which $\sigma$ is non-vanishing is open.
Equivalently, write the trivialised forms of $\sigma_1 \dots \sigma_n$ as a big $r \times n$ matrix of smooth functions, where $r$ is the rank of the vector bundle. Saying that $\sigma_1(p) \dots \sigma_n(p)$ are linearly independent is the same as saying that the determinant of at least one $n \times n$ submatrix within your big matrix is non-vanishing at $p$. Then use continuity of these determinants.


TITLE: Showing that $|\cos(x)|+|\sin(x)|\le \sqrt{2}$
QUESTION [1 upvotes]: Show that for all $x\in\mathbb{R}$, $|\cos(x)|+|\sin(x)|\le \sqrt{2}$

My problem is mainly with the modules. I proved it for $f(x) = \cos(x) + \sin(x)$ but don't know how to take this to the modules case. Thanks in advance.
For $f(x)$: take $f'(x) = 0 \iff -\sin(x) + \cos(x) = 0 \iff \cos(x) = \sin(x) \iff x=\frac{\pi}{4}+k\pi \Rightarrow f(x)\le \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}$

REPLY [5 votes]: $(x+y)^2 \le 2(x^2+y^2)$.
Put $x=|\cos x|, y=|\sin x|$.


TITLE: $\int \frac{1}{(x+2)^{1/3}(x-1)^{2/3}}dx$
QUESTION [0 upvotes]: How can I do the following integral?
$\int \frac{1}{(x+2)^{1/3}(x-1)^{2/3}}dx$

I have tried a lot of methods but I don't know how to do it

REPLY [1 votes]: I solve this for you, though it acquires many steps of substitution and doing partial fractions. There may be some mistakes. Let me know if you have any questions or have found any typos or mistakes. I skip some algebra. You can complete on your own.
First, substitute $u=(x-1)^{1/3}$, $*=3\int\frac{1}{{u^3+3}^{1/3}}du$.
Second, substitute $v=\frac{{u^3+3}^{1/3}}{u}$ and apply partial fraction, $*=-\int\frac{v}{(v-1)(v^2+v+1)}dv=\frac{1}{3}\int\frac{1}{v-1}dv-\frac{1}{3}\int\frac{v-1}{v^2+v+1}dv$. The first integral is $\ln(v-1)$, and the second one, is $\frac{1}{2}\int\frac{2v+1}{v^2+v+1}dv-\frac{3}{2}\int\frac{1}{v^2+v+1}dv$.
The first integral is $\ln(v^2+v+1)$, and for the second one, complete squares on the denominator and substitute $w=\frac{2v+1}{\sqrt{3}}$, get $=\frac{2\arctan(w)}{\sqrt{3}}$.
The trick is now over, and substitute every dummy variable back, and by some algebra, the answer is
$=\frac{\ln\big(\frac{(x+2)^{2/3}}{(x-1)^{2/3}}+\frac{(x+2)^{1/3}}{(x-1)^{1/3}}+1\big)}{2}-\sqrt{3}\arctan\big(\frac{\frac{2(x+2)^{1/3}}{(x-1)^{1/3}}+1}{\sqrt{3}}\big)-\ln(\frac{(x+2)^{1/3}}{(x-1)^{1/3}}-1)$.
If you find this helpful, please vote.


text:
\begin{definition}[Definition:Fixed Element of Permutation/Moved]
$x$ '''moved by $\pi$''' {{iff}}:
:$\map \pi x \ne x$
\end{definition}

config: wiki


TITLE: show that $L_{0}(x)=1$ ,$L_{1}(x)=1-x$ and $L_{2}(x)=1-2x +\frac{x^{2}}{2}$ are mutually orthogonal
QUESTION [0 upvotes]: show that $L_{0}(x)=1$ ,$L_{1}(x)=1-x$ and $L_{2}(x)=1-2x +\frac{x^{2}}{2}$ are mutually orthogonal on the interval $(0,\infty )$ with respect to the weighting function $r(x)=e^{-x}$
I have trouble advancing with this problem, though it seemed quite straight forward at glance.
I've done this, but I don't know if it's okay
$\left \langle L_{0},L_{1} \right \rangle=\int_{0}^{\infty }(1-x)e^{-x}dx$
$\left \langle L_{0},L_{2} \right \rangle=\int_{0}^{\infty }(1-2x +\frac{x^{2}}{2})e^{-x}dx$
$\left \langle L_{1},L_{2} \right \rangle=\int_{0}^{\infty }(1-x)(1-2x +\frac{x^{2}}{2})e^{-x}dx$

REPLY [1 votes]: Yes, your approach is right. Now all you need is to compute the integrals. Note that you can simplify your calculations. After you show $\left \langle L_{0},L_{1} \right \rangle=0$, you rewrite $\left \langle L_{0},L_{2} \right \rangle$ as $$\left \langle L_{0},L_{2} \right \rangle=\int_0^\infty(1-x)e^{-x}dx+\int_0^\infty(-x+\frac{x^2}{2})e^{-x}dx$$
The first term is just $\left \langle L_{0},L_{1} \right \rangle=0$. Similarly, for the last integral you write $$\left \langle L_{1},L_{2} \right \rangle=\int_0^\infty(1-2x+\frac{x^2}{2})e^{-x}dx+\int_0^\infty(-x+2x^2-\frac{x^3}{2})e^{-x}dx\\=\left \langle L_{0},L_{2} \right \rangle+\int_0^\infty(-x+\frac{x^2}{2}+\frac{3x^2}{2}-\frac{x^3}{2})e^{-x}dx$$
Notice that the last integral you already computed the integral of the first two terms.


TITLE: Suppose that $x$ and $y$ satisfy $\frac{x}{2} + \frac{y}{3} = 1$. Prove that $x^2 + y^2 > 1$.
QUESTION [1 upvotes]: Ok , i tried to prove this via Contrapositive setting $x^2 + y^2 \le 1$. After doing some algebra i have arrived at $x \le \sqrt{-y^2}$. I'm fairly sure this isn't right. I also solved for x and y in equation one hoping this would somehow lead me to a conclusion, it didn't.

REPLY [0 votes]: Here are two geometric ways to prove this:
Method.$(1)$
Let the line intersect with the x-axis at point $A(2,0)$, and intersect with the y-axis at point $B(0,3)$. Draw the height from  $O$ to the hypotenuse and intersect with $AB$ at $H$, $\Rightarrow OH\perp AB$. So $OH$ is the shortest distance from $O$ to the straight line $AB$.
So we have $OA=2,~OB=3,~AB=\sqrt{13}$. From the triangle area, we have:
$$\frac{1}{2}\cdot OA\cdot OB=\frac{1}{2}\cdot AB\cdot OH~~\Rightarrow~~OH=\frac{6}{\sqrt{13}}$$
Because $OH$ is the shortest distance from $O$ to the straight line $AB$, and $OH>1=R$, where the radius of the circle is $R=1$. Therefore, the straight line $AB$ is outside the circle. So for all points on the straight line, which satisfying $\frac{x}2+\frac{y}3=1$, their distances to the origin $O$ are greater than $1$. Namely, $x^2+y^2>1$.
Method.$(2)$
Combine two equations:
$$\frac{x}2+\frac{y}3=1,~~~x^2+y^2=1$$
Eliminating $x$ (or eliminating $y$), and we get a quadratic equation, it is easy to show it has no real roots, which means there are no intersecting points or touching points between the straight line and the circle. This means the line is outside the circle. So for all points on the straight line, which satisfying $\frac{x}2+\frac{y}3=1$, their distances to the origin $O$ are greater than $1$. Namely, $x^2+y^2>1$.


TITLE: Counting techniques to improve efficiency
QUESTION [1 upvotes]: I am having trouble with the following question:

There are four people named W,X,Y,Z. Each person is currently in a room and each room can only hold one person. If we randomly shuffle these four people between different rooms, what is the probability that exactly two of these people will end up in the same room where they started in the beginning.

I let the event $A$ represent two people returning to the same rooms. I let $n(A)$ be the number of ways this event can happen. I am able to calculate $n(A)$ but I am looking for a more efficient way to do so.
I let $n(A) = (1\times3\times1\times1) + (3\times1\times1\times1)$
since I realized that the outcomes would depend on whether the first person entered the correct room. If the number of people and rooms increased this method would take too long.

REPLY [1 votes]: In order for exactly two of the people to end up in their original rooms, the shuffle must simply have switched the other two people. For instance, if $W$ and $X$ end up in their original rooms, $Y$ and $Z$ have simply switched rooms. There’s exactly one shuffle that leaves $W$ and $X$ as the only people in their original rooms. Thus, all you have to do is note that there are $\binom42=6$ pairs of people: for each pair there is one shuffle leaving that pair fixed and interchanging the other pair.
This is the most efficient way when there are $4$ people, but it won’t generalize. The general problem is a bit nasty. Suppose that you have $n$ people. There are $\binom{n}2$ ways to choose the $2$ people who get their original rooms. The shuffle must then move each of the other $n-2$ people to a different room. In other words, you need a derangement of the other $n-2$ people. The link gives formulas for the number of derangements; perhaps the nicest from a purely computational point of view is that the number of derangements of a set of $n$ things is
$$\left[\frac{n!}e\right]=\left\lfloor\frac{n!}e+\frac12\right\rfloor\;,$$
where $[x]$ is the integer nearest $x$. The article also gives a couple of useful recurrences, and there is more information, with references, in the OEIS entry for the sequence of derangement numbers.


text:
\begin{document}

\pagestyle{plain}

\maketitle

\paragraph{Abstract\,:} We introduce the notion of a Thom class of a current and define the localized intersection of currents.
In particular we consider the situation where we have a $C^\infty$ map of \mfd s and study  localized
intersections of the source \mfd\ and currents on the target \mfd. We then obtain a residue theorem
on the source \mfd\ and give  explicit formulas for the residues in some cases.
These are applied to the problem of coincidence points of two maps.
We define the global and local coincidence homology classes and indices.
A representation of the Thom class of the graph as a \v Cech-de~Rham cocycle immediately gives us an explicit expression of the  index at an isolated coincidence point, which in turn gives explicit coincidence classes in some non-isolated
components. Combining these, we have a general coincidence point theorem including the one by S. Lefschetz.

\vv

\noindent
{\it Keywords}\,: Alexander duality, Thom class, localized intersections, residue theorem, coincidence
classes and indices, Lefschetz coincidence point formula.

\vv

\noindent
{\it Mathematics Subject Classification} (2010)\,: 14C17, 32C30, 37C25, 55M05, 55M20,
57R20, 58A25.

\section*{Introduction}

For two cycles in a \mfd, the localized intersection product of their classes
is defined,   in the homology of their set theoretical intersection, via Alexander dualities and the cup product in the relative
cohomology.
Thus for a cycle $C$ in a \mfd\ $W$, the corresponding class   in the relative cohomology carries the local
information on $C$. For a sub\mfd\ $M$ of $W$, this is the Thom class of $M$, which may be identified with
the Thom class of the normal bundle of $M$ in $W$ by the tubular \nbd\ theorem.
In this paper we take up the localization problem of currents. We introduce the notion of a Thom class of a current
and define the localized intersection of two currents. In particular we consider the intersections of a fixed sub\mfd\ $M$
and currents on $W$, obtain a residue theorem on $M$ and give explicit expressions of the residues in some cases
(see Theorems \ref{res-m} and \ref{resnoniso} below for precise statements).
As an application we study the coincidence point problem for two maps.

The coincidence point formula discovered by S. Lefschetz (cf. \cite{Lef1}, \cite{Lef2}) is   formulated for a pair of continuous maps
between compact oriented  topological manifolds of the same dimension.
Using the above,  we define global and local homology classes of
coincidence for a pair of $C^\infty$ maps $M\ra N$ and give a general coincidence point theorem,
even in the case  the dimensions $m$ and $n$ of $M$ and $N$ are different ($m\ge n$) and
the coincidence points are non-isolated (Definitions \ref{gcoin} and \ref{lcoin}). In the case  $m=n$,
the use of Thom class in the \v Cech-de~Rham cohomology immediately gives us an explicit
expression of the
coincidence index at an isolated  coincidence point (Propositions \ref{propisolated}). This gives in turn an explicit coincidence homology class
at a certain non-isolated coincidence component  in the case $m>n$ (Proposition \ref{coinnonisol}).
We then have a general
coincidence point formula, including the Lefschetz coincidence point formula.

The paper is organized as follows. In Section \ref{CdRTh}, we recall preliminary materials such as  Poincar\'e and Alexander dualities, global and localized intersection products,  Thom classes
in various settings and an explicit expression, in the \v Cech-de Rham cohomology, of the Thom class of an oriented \vb\
(Proposition \ref{Thomloc}). In Section \ref{secloc}, we consider the localization problem of currents, introduce the notion of
a Thom class of a current and give some examples. We then consider localized intersections of currents.
In particular we study  intersections of a fixed sub\mfd\ $M$ in a \mfd\ $W$ and currents on $W$.
In fact we consider a more general situation where we have a map $F:M\ra W$ (Definition \ref{prodmap}).
 For a closed current $T$ on $W$, we have the intersection product $M\cdot_F T$ in the homology of $M$ and, if $T$ is localized at a compact set $\tilde S$ in $W$
and if $\vP_T$ is a Thom class of $T$ along $\tilde S$, we have the residue of $F^*\vP_T$  in the homology of $S=F^{-1}\tilde S$ as a localized intersection product. If $S$ has several connected components, we have a ``residue
theorem" (Theorem \ref{res-m}). We give an explicit formula for the residue at a non-isolated component of $S$ in the case it
is a sub\mfd\ of $M$ (Theorem \ref{resnoniso}). These are  conveniently used in Section \ref{seccoin}, where we study
the coincidence point problem for two maps.

Let $M$ and $N$ be  \mfd s of dimensions $m$ and $n$ and let
$f, g:M\ra N$ be two maps.  We define the global coincidence class of the pair $(f,g)$ in the $(m-n)$-th homology of $M$
(Definition \ref{gcoin}). If $M$ is compact, we also define the local coincidence class of the pair  in the $(m-n)$-th homology of the  set of points in $M$ where $f$ and $g$ coincide (Definition \ref{lcoin}). We then
apply Theorem \ref{res-m} to get a general coincidence point theorem (Theorem \ref{gcth}).
In the case $m=n$ we have a formula for the coincidence index at an isolated coincidence point as the
local mapping degree of $g-f$ (Proposition \ref{propisolated}).
This is in fact a classical result, however we give a short direct proof using the aforementioned expression of the Thom
class in the \v Cech-de Rham cohomology. This together with Theorem \ref{resnoniso} gives an explicit expression of the
coincidence homology class at a non-isolated component (Proposition \ref{coinnonisol}). If $M$ and $N$ are compact
\mfd s of the same dimension, we have a general coincidence point formula (Theorem \ref{non-isolatecoin}), which
reduces to the Lefschetz coincidence point formula in the case the coincidence points are isolated (Corollary \ref{lefcpf}).

\section{Notation and conventions}\label{secnote}

For a topological space $X$, we  denote by $H_*(X,\C)$ and $H^*(X,\C)$ its homology and cohomology
 of finite singular chains with $\C$ coefficients. Also we denote by $\breve H_*(X,\C)$  the homology of locally finite
 singular chains (Borel-Moore homology). For a $C^\infty$ \mfd, they can be computed using $C^\infty$
 simplicial chains. To be a little more precise, any $C^\infty$ \mfd\ $M$ admits a $C^\infty$ triangulation, which is essentially unique,

and the groups $H_*(M,\C)$ and $H^*(M,\C)$ are naturally isomorphic with the ones defined by
  finite $C^\infty$ simplicial chains. Also the group $\breve H_*(M,\C)$  is naturally isomorphic with the homology of locally finite $C^\infty$ simplicial chains.

 In the sequel  a locally finite $C^\infty$ simplicial chain is simply called a chain,
unless otherwise  stated. Thus a chain $C$ is expressed as a locally finite sum $C=\sum a_is_i$ with  $a_i$  in $\C$ and $s_i$
 oriented $C^\infty$ simplices. We set $|C|=\bigcup s_i$ and call it the support of $C$. It is a closed set.
 For a cycle $C$, its class in the homology of the ambient space  is denoted by $[C]$, while the class in the homology
 of its support is simply denoted by $C$.

For an open set $U$ in $M$, we
denote by $A^p(U)$ and $A^p_c(U)$, \r, the spaces of complex valued $C^{\infty}$ $p$-forms and $p$-forms with
compact support on $U$. The
cohomology of the complex $(A^*(M),d)$ is the de~Rham cohomology
$H^*_{\rm dR}(M)$ and that of $(A^*_c(M),d)$ is the cohomology $H^*_c(M)$ with compact support.
A $C^\infty$ form will be  simply called a  form unless otherwise  stated.

\section{\v Cech-de~Rham cohomology and the Thom class}\label{CdRTh}

For the background on the \v Cech-de~Rham cohomology, we refer to \cite{BT}. The integration theory
on this cohomology  is developed in  \cite{Leh2}.
See  \cite{Su6} also for these materials and for the description of the Thom class in the framework of  relative \v Cech-de~Rham cohomology. The relation with the combinatorial viewpoint, as given in \cite{Br1},  is discussed in \cite{Su8}.

In this section we let $M$ denote a $C^\infty$ \mfd\ of dimension $m$.

\subsection{Poincar\'e duality}\label{ssecPoin}

We recall the Poincar\'e duality  and global intersection products of homology classes.

Suppose that $M$ is connected and oriented. Then  the pairing
\[
A^p(M)\times A^{m-p}_c(M)\lra\C\qquad\text{given by}\quad (\o,\varphi)\mapsto\int_M\o\wedge\varphi
\]
induces the Poincar\'e duality for a possibly non-compact \mfd\,:
\begin{equationth}\label{PoinDual}
P:H^p(M,\C)\simeq H^p_{\rm dR}(M)\stackrel{\sim}{\lra} H^{m-p}_c(M)^*\simeq\breve H_{m-p}(M,\C).
\end{equationth}
In the sequel we sometimes omit the coefficient $\C$ in homology and cohomology. In fact the Poincar\'e duality holds with $\Z$ coefficient. Note that $P$ is given by the left cap product with the fundamental class
of $M$, the class of the sum of all $m$-simplices in $M$.
We also denote $P$ by $P_M$ if we wish  to make the \mfd\ $M$ under consideration explicit.

In the  \iso\ \eqref{PoinDual}, the class $[\o]$ of a closed $p$-form $\o$ corresponds

to the functional
on $H^{m-p}_c(M)$ given by
\begin{equationth}\label{corrpoinform}
[\varphi]\mapsto\int_M\o\wedge\varphi,
\end{equationth}
or

to the class $[C]$ of an $(m-p)$-cycle $C$ \st
\begin{equationth}\label{corrpoin}
\int_M\o\wedge\varphi=\int_C\varphi
\end{equationth}
for any closed form $\varphi$ in $A^{m-p}_c(M)$. We call $\o$ a {\em  de~Rham representative} of $C$.

For
two classes $[C_1]\in \breve H_{q_1}(M)$ and $[C_2]\in \breve H_{q_2}(M)$,
the {\em intersection product} $[C_1]\cdot [C_2]$ is
defined by
\begin{equationth}\label{defint}
[C_1]\cdot [C_2]:=P(P^{-1}[C_1]\smallsmile P^{-1}[C_2])\qquad\text{in}\ \ \breve H_{q_1+q_2-m}(M),
\end{equationth}
where $\smallsmile$ denotes the cup product, which corresponds to the exterior product in the first \iso\ in \eqref{PoinDual}.

If $M$ is compact and connected, then $\breve H_0(M,\C)=H_0(M,\C)=\C$. Thus if $q_1+q_2=m$, $[C_1]\cdot [C_2]$ is a number given by
\[
[C_1]\cdot [C_2]=\int_M\o_1\wedge\o_2=\int_{C_1}\o_2=(-1)^{q_1q_2}\int_{C_2}\o_1,
\]
where $\o_1$ and $\o_2$ are de~Rham representatives of $C_1$ and $C_2$, \r.

\subsection{\v Cech-de~Rham cohomology}\label{CdR}

The \v Cech-de~Rham cohomology is defined for an arbitrary open covering of
$M$, however here we only consider  coverings  consisting of  two open sets.
Thus let $\U=\{U_0, U_1\}$ be an open covering of $M$. We set
$U_{01}=U_0\cap U_1$ and define the complex vector space $A^p(\U)$ as
\[
A^p(\U):=A^p(U_0)\oplus A^p(U_1)\oplus A^{p-1}(U_{01}).
\]
An element $\sigma$ in $A^p(\U)$ is given by a triple
$\sigma=(\sigma_0, \sigma_1, \sigma_{01})$ with $\sigma_i$ a $p$-form on $U_i$, $i=0,1$, and
$\sigma_{01}$  a $(p-1)$-form on $U_{01}$. We define an operator $D: A^p(\U) \to A^{p+1}(\U)$ by
\[
D\sigma:=(d\sigma_0, d\sigma_1, \sigma_1|_{U_{01}}-\sigma_0|_{U_{01}}-d\sigma_{01}).
\]
Then we see that  $D \circ D =0$ so that we have a complex
$(A^*(\U),D)$. The $p$-th {\em \v{C}ech-de~Rham cohomology} of $\U$, denoted by
$H_D^p(\U)$, is the $p$-th cohomology of this complex. It is also abbreviated as  \v CdR cohomology.
We denote the class of a cocycle $\sigma$ by $[\sigma]$.
It can be shown that the map $A^p(M) \to A^p(\U)$ given by
$\o \mapsto (\o|_{U_0}, \o|_{U_1}, 0)$
induces an isomorphism
\begin{equationth}\label{dRCdR}
\a:H^p_{\rm dR}(M)\stackrel\sim\lra  H^p_D(\U).
\end{equationth}
Note that $\a^{-1}$ assigns to the class of a \v CdR cocycle $(\sigma_0,\sigma_1,\sigma_{01})$ the class of the
closed form $\rho_0\sigma_0+\rho_1\sigma_1-d\rho_0\wedge\sigma_{01}$, where $\{\rho_0,\rho_1\}$ is a partition
of unity subordinate to $\U$.
\vv

Now we could define the cup product for \v CdR cochains and describe the Poincar\'e duality
in terms of the \v CdR cohomology as in \cite{Su6} in the case $M$ is compact.  However here we proceed as follows.
Let $M$  and $\U=\{U_0, U_1\}$ be as above. A  {\em system of honeycomb cells} adapted to $\U$ is a collection
$\{R_0,R_1\}$
of two  submanifolds of $M$ of dimension $m$ with $C^\infty$ boundary
having the following properties\,:
\smallskip

\noindent
(1) $R_i\subset U_i$ for $i=0,1$,
\smallskip

\noindent
(2) $\op{Int}\, R_0\cap \op{Int}\, R_1=\emptyset$ and
\smallskip

\noindent
(3)  $R_0 \cup R_1=M$,
\smallskip

\noindent
where $\op{Int}$ denotes the interior.
Suppose $M$ is oriented. Then $R_0$ and $R_1$ are naturally oriented. Let $R_{01}=R_0\cap R_1$ with  the orientation as the boundary of
$R_0$\,; $R_{01}=\partial R_0$, or equivalently,  the  orientation opposite to that of  the
boundary of $R_1$\,; $R_{01}=-\partial R_1$. We consider the pairing
\[
A^p(\U)\times A^{m-p}_c(M)\lra\C
\]
given by
\begin{equationth}\label{intCdR}
(\sigma,\varphi)\mapsto\int_{R_0}\sigma_0\wedge\varphi+\int_{R_1}\sigma_1\wedge\varphi+\int_{R_{01}}\sigma_{01}\wedge\varphi.
\end{equationth}
Then it induces the Poincar\'e duality \eqref{PoinDual} through the \iso\ $\a$ in \eqref{dRCdR}.

\subsection{Relative \v Cech-de~Rham cohomology and Alexander duality}\label{relCdR}

We introduce the relative \v Cech-de~Rham cohomology and describe the Alexander duality, which is used to define
localized intersection products.

Let  $S$ be a closed set in $M$. Letting
$U_0=M\ssm  S$ and $U_1$ a \nbd\ of $S$ in $M$, we
consider the covering $\U=\{U_0,U_1\}$ of $M$.
 If we set
\[
A^p(\U,U_0)=\{\,\sigma\in A^p(\U)\mid \sigma_0=0\,\},
\]
we see that $(A^*(\U,U_0),D)$ is a subcomplex of $(A^*(\U),D)$. We denote by
$H^p_D(\U,U_0)$ the $p$-th cohomology of this complex. From the short exact sequence
\[
0\lra A^*(\U,U_0)\stackrel{j^*}\lra A^*(\U)\stackrel{\iota^*}\lra A^*(U_0)\lra 0,
\]
where $j^*$ is the inclusion and $\iota^*$ is the \homo\ that assigns $\sigma_0$ to $\sigma=(\sigma_0,\sigma_1,\sigma_{01})$, we have the long exact sequence
\begin{equationth}\label{longexact}
\cdots\lra H^{p-1}_D(\U)\stackrel{\iota^*}\lra H^{p-1}_{\rm dR}(U_0)\stackrel{\delta^*}\lra H^p_D(\U,U_0)\stackrel{j^*}\lra H^p_D(\U)\stackrel{\iota^*}\lra H^p_{\rm dR}(U_0)\lra\cdots.
\end{equationth}
In the above, $\delta^*$ assigns the  class $[(0,0,-\t)]$ to the class of a closed $(p-1)$-form $\t$ on $U_0$.
Comparing with the long cohomology exact sequence for the pair $(M,M\ssm S)$, we have
 a natural \iso\ (see  \cite{Su8} for a precise proof)\,:
\[
H^p_D(\U,U_0)\simeq H^p(M,M\ssm S;\C).
\]

We  describe the Alexander duality in terms of the relative \v CdR cohomology  in the case $S$ is compact
and admits a regular \nbd\ (cf.  \cite{Su6}).
Thus suppose  $M$ is oriented and let $\{R_0,R_1\}$ be a system of honeycomb cells adapted to $\U$.
We assume that $S$ is compact so that we may also assume that $R_1$ is compact. Consider the pairing
\begin{equationth}\label{relpair}
A^p(\U,U_0)\times A^{m-p}(U_1)\lra\C
\end{equationth}
given by
\[
(\sigma,\varphi)\mapsto\int_{R_1}\sigma_1\wedge\varphi+\int_{R_{01}}\sigma_{01}\wedge\varphi.
\]
Then it induces the Alexander \homo\,:
\begin{equationth}\label{alexhomo}
A:H^p(M,M\ssm S;\C)\simeq H^p_D(\U,U_0)\lra H^{m-p}_{\rm dR}(U_1)^*\simeq H_{m-p}(U_1,\C).
\end{equationth}
The \homo\ \eqref{alexhomo}
 depends on $U_1$ and is not an \iso\ in general.
Here we consider the following hypothesis\,:
\begin{enumerate}
\item[(*)] there exists a triangulation of $M$ \st\
$S$ is (the polyhedron of) a subcomplex.
\end{enumerate}
In this case we also say that the triangulation is compatible with $S$.

\begin{remark} The triangulation above may simply be $C^0$ to have the Alexander duality
\eqref{AlexDual} below. We mainly consider $C^\infty$ triangulations in order that, in an  expression as \eqref{corralex}, the integral in the right hand side  makes
sense. The hypothesis is certainly satisfied  if $S$ is the support of a chain.
It is also satisfied if $S$ is a $C^\infty$
sub\mfd\ of $M$.

Another  case of interest where the following arguments go through is that $S$ is a subanalytic set in a real analytic \mfd\ $M$, as in this case there is a subanalytic triangulation of $M$ compatible with $S$ (cf. \cite{Sh}) so that the integration makes sense, see Example \ref{exmult} below.
\end{remark}

Under the hypothesis (*), we may take as $U_1$ a regular \nbd\ of $S$ so that there is a
deformation retract  $U_1\ra S$.  We then have $H_{m-p}(U_1)\simeq H_{m-p}(S)$ and
\eqref{relpair} induces the Alexander duality\,:

\begin{equationth}\label{AlexDual}
A:H^p(M,M\ssm S;\C)\simeq H^p_D(\U,U_0)\stackrel{\sim}{\lra} H^{m-p}_{\rm dR}(U_1)^*\simeq H_{m-p}(S,\C).
\end{equationth}
Note that $A$ is also given by the left cap product with the fundamental class $M$. We also denote $A$ by $A_{M,S}$ if we wish  to make the pair $(M,S)$ explicit.

In the  \iso\ \eqref{AlexDual}, the class $[\sigma]$ of a  $p$-cochain $\sigma$ corresponds to the functional
on $H^{m-p}_{\rm dR}(U_1)$ given by
\begin{equationth}\label{corralexform}
[\varphi_1]\mapsto\int_{R_1}\sigma_1\wedge\varphi_1+\int_{R_{01}}\sigma_{01}\wedge\varphi_1,
\end{equationth}
or to the class $[C]$ of an $(m-p)$-cycle $C$  in $S$ \st
\begin{equationth}\label{corralex}
\int_{R_1}\sigma_1\wedge\varphi_1+\int_{R_{01}}\sigma_{01}\wedge\varphi_1=\int_C\varphi_1
\end{equationth}
for any closed form $\varphi_1$ in $A^{m-p}(U_1)$.

Denoting by $i:S\hra M$ the inclusion, we have the following commutative diagram\,:
\begin{equationth}\label{PoinAlex}
\SelectTips{cm}{}
\xymatrix{ H^p(M, M \ssm S)\ar[r]^-{\sim}_-A \ar[d]^{j^*}&H^{m-p}_{\rm dR}(U_1)^*\simeq  H_{m-p}(S)
\ar[d]^{i_*}\\
H^p(M) \ar[r]^-{\sim}_-P &H^{m-p}_c(M)^*\simeq \breve H_{m-p}(M).}
\end{equationth}

\begin{remark}\label{remimp}
1.
In the above, the \homo\ $i_*:H_{m-p}(S)\ra \breve H_{m-p}(M)$ is the one naturally induced from $i$, while
$i_*:H^{m-p}_{\rm dR}(U_1)^*\ra H^{m-p}_c(M)^*$
is described as follows.
For any functional $F_1$ on $H^{m-p}_{\rm dR}(U_1)$, there is a corresponding cycle $C$ in $S$ and $i_*F_1$ is given by
\[
i_*F_1[\varphi]=\int_C\varphi\qquad\text{for}\ \ [\varphi]\in H^{m-p}_c(M).
\]
Alternatively, if $(0,\sigma_1,\sigma_{01})$ is a \v CdR representative of $A^{-1}F_1$, then
\[
i_*F_1[\varphi]=\int_{R_1}\sigma_1\wedge\varphi+\int_{R_{01}}\sigma_{01}\wedge\varphi.
\]
\smallskip

\noindent
2. For a closed set $S$ (which may not be compact) in $M$ satisfying (*), we may define the Alexander \iso
\[
A:H^p(M,M\ssm S)\stackrel{\sim}{\lra} \breve H_{m-p}(S)
\]
via combinatorial topology (cf. \cite{Br1}).
\end{remark}

Let $S_1$ and $S_2$ be compact sets in $M$ satisfying (*) and set $S=S_1\cap S_2$. Let $A_1$, $A_2$ and $A$ denote the Alexander \iso s for $(M,S_1)$, $(M,S_2)$ and $(M,S)$, \r.
For two classes $c_1\in  H_{q_1}(S_1)$ and $c_2\in  H_{q_2}(S_2)$, the {\em localized intersection
product} $(c_1\cdot c_2)_S$ is defined by
\begin{equationth}\label{locint}
(c_1\cdot c_2)_S:=A(A_1^{-1}c_1\smallsmile A_2^{-1}c_2)\qquad\text{in}\ \  H_{q_1+q_2-m}(S),
\end{equationth}
where $\smallsmile$ denotes the cup product
\[
H^{m-q_1}(M,M\ssm S_1)\times H^{m-q_2}(M,M\ssm S_2)\stackrel\smallsmile\lra H^{2m-q_1-q_2}(M,M\ssm S).
\]
Letting $i_1:S_1\hra M$, $i_2:S_2\hra M$ and $i:S\hra M$ be the inclusions, from
\eqref{PoinAlex} we see that the definitions \eqref{defint} and \eqref{locint} are consistent in the sense that
\[
i_*(c_1\cdot c_2)_S=(i_1)_*c_1\cdot (i_2)_*c_2.
\]

\subsection{Thom class}\label{Thom}

We list \cite{MS} as a general reference for the Thom \iso\ and the Thom class of a  real \vb. In general
they are defined in cohomology with $\Z_2$ coefficients, while
for an oriented \vb, they can be defined in cohomology with $\Z$ coefficients.
They can also be described in terms of differential forms,
in which case the cohomology involved is with $\C$ coefficients. This is done
in \cite{BT} using cohomology with compact support in the vertical direction.
Here we use \v Cech-de~Rham cohomology instead as in \cite{Su6}. This way we can express relevant local informations
more explicitly.

In this subsection, we sometimes omit the coefficient $\C$ in homology and cohomology.
In fact the \iso s we consider below can be defined from combinatorial viewpoint in homology and cohomology with  $\Z$ coefficient (cf. \cite{Br1}, also \cite{Su8}).

\paragraph{(a) Thom class of an oriented real \vb\,:} Let
$\pi:E\to M$ be an oriented real \vb\ of rank $k$. We identify $M$ with the image of the zero section. Then we have the Thom \iso\
\[
T_E:H^p(M,\C)\overset{\sim}{\lra}H^{p+k}(E,E\ssm M;\C),
\]
 whose inverse is given by the integration   along the fiber   of $\pi$ (see \cite[Ch.II, 5]{Su6}).

 The {\em Thom class
 $\vP_E$} of $E$, which is in $H^{k}(E,E\ssm M)$, is the image of the constant function $1$ in $H^0(M,\C)$ by $T_E$.
 Note that $T_E$ is given by the cup product with $\vP_E$. Let $W_0= E\ssm M$ and $W_1$ a \nbd\ of $M$ in $E$ and consider the covering $\W=\{W_0,W_1\}$ of  $E$. We refer to \cite[Ch.II, Proposition 5.7]{Su6} for an explicit
 expression of a \v CdR cocycle representing $\vP_E$
in the \iso\ $H^k(E,E\ssm M)\simeq H^k_D(\W,W_0)$.
In particular suppose $E$ is trivial on an open set $U$ of $M$. Then, setting $A^k(\W,W_0)|_U=A^k(\W',W_0')$ with
$W_i'=W_i\cap\pi^{-1}(U)$, we have (cf. \cite[Ch.III, Lemma 1.4]{Su6})\,:

\begin{proposition}\label{Thomloc}
Suppose $E$ is trivial on an open set $U$ of $M$\,; $E|_U\simeq\R^k\times U$ and let $\rho:E|_U\ra\R^k$ denote the
projection on to the fiber direction. Then $\vP_{E|_U}$ is represented by a cocycle in $A^k(\W,W_0)|_U$ of the form
\[
(0,0,-\rho^*\psi_k),
\]
where $\psi_k$ is an angular form on $\R^k\ssm\{0\}$, i.e., a closed $(k-1)$-form with $\int_{S^{k-1}}\psi_k=1$.
\end{proposition}

Suppose $M$ is compact and oriented. We orient the total space $E$ so that, if $\xi=(\xi_1,\dots,\xi_k)$ is a positive fiber coordinate system of $E$ and if $x=(x_1,\dots,x_m)$ is a positive coordinate system on $M$, then $(\xi,x)$ is a positive coordinate system on $E$.
We then have the commutative diagram\,:

\begin{equationth}\label{ThAP}
\SelectTips{cm}{}
\xymatrix{H^p(M) \ar[r]_-{T_E}^-{\sim} \ar[d]^{P}_{\wr}& H^{p+k}(E,E\ssm M)
\ar[ld]^{A}_{\rotatebox{30}{{\hspace{-3.5mm}}$\sim$}}  \\
H_{m-p}(M).&{}}
\end{equationth}

\paragraph{(b) Thom class of a sub\mfd\,:}
Let $W$ be an oriented $C^\infty$ \mfd\ of dimension $m'$ and $M$ a compact and
oriented sub\mfd\ of $W$ of dimension $m$. Set $k=m'-m$.
In view of (\ref{ThAP}), we define the Thom \iso\ $T_M:H^p(M)\overset{\sim}\ra H^{p+k}(W,W\ssm M)$
so that the following  diagram becomes commutative (cf. \cite{Br1})\,:
\[
\SelectTips{cm}{}
\xymatrix{H^p(M) \ar[r]_-{T_M}^-{\sim} \ar[d]^{P}_{\wr}& H^{p+k}(W,W\ssm M)
\ar[ld]^{A}_{\rotatebox{30}{{\hspace{-3.5mm}}$\sim$}}  \\
H_{m-p}(M).&{}}
\]
Then  the Thom class $\vP_M$ of $M$
is defined by
\begin{equationth} \label{ThomClass}
\vP_M=:T_M(1)=A^{-1}(M)\qquad\text{in}\ \ H^{k}(W,W\ssm M).
\end{equationth}

Let $N_M\to M$ denote the normal bundle of $M$ in $W$. Suppose the orientation of $M$ is compatible with that of $W$
in the sense that $N_M$ is orientable.
We orient $N_M$ as follows. Namely, if $(x_1,\dots,x_k,\dots,x_{m'})$ is a positive coordinate system on $W$ such that $M$ is given by $x_1=\cdots=x_k=0$ and that $(x_{k+1},\dots,x_{m'})$ is a positive coordinate system of $M$, then the vectors $(\frac{\partial}{\partial x_1},\dots,\frac{\partial}{\partial x_k})$ determine a positive frame of $N_M$. By the tubular \nbd\ theorem, there is a \nbd\ $W_1$ of $M$ in $W$ and an orientation
preserving diffeomorphism of $W_1$ onto a \nbd\ of the zero section of $N_M$, which is identified with $M$.
By excision we have
\[
H^k(W,W\ssm M)\simeq
H^k(W_1,W_1\ssm M)\simeq H^k(N_M,N_M\ssm M)
\]
and, in the above \iso s, the Thom class $\vP_M$ of $M$ corresponds to the Thom class $\vP_{N_M}$ of the \vb\ $N_M$.

\begin{remark}\label{rempseudo} More generally, for a pseudo-\mfd\ $M$ in $W$, we may define the Poincar\'e \homo\
$P:H^p(M)\lra\breve H_{m-p}(M)$.
Thus we have the Thom \homo\ and the Thom class $\vP_M$ of $M$ (cf.  \cite{Br1}).
\end{remark}
\paragraph{(c) Thom class of a cycle\,:} Let $C$ be a finite $(m'-p)$-cycle in $W$ and $\tilde S$ its support.
We may define the Thom class $\vP_C$ of $C$ by
\[
\vP_C:=A^{-1}(C),
\]
where $A$ is the Alexander \iso\ \eqref{AlexDual} for the pair $(W,\tilde S)$.

\section{Localized intersection of currents}\label{secloc}

\subsection{Thom class of a current}\label{thomclassofcurrents}

Let $W$ be an oriented  $C^\infty$ manifold of dimension $m'$. Recall that a  {\em $p$-current} $T$ on $W$ is a continuous linear functional
on the space $A_c^{m'-p}(W)$. We use the notation
\[
T(\varphi)=\langle T,\varphi\rangle,\qquad\varphi\in A_c^{m'-p}(W).
\]
Let $\D^p(W)$ denote the space of $p$-currents on $W$. The differential $d:\D^p(W)\to \D^{p+1}(W)$ is defined by
\[
\langle dT,\varphi\rangle=(-1)^{p+1}\langle T,d\varphi\rangle,\qquad\varphi\in A_c^{m'-p-1}(W).
\]
Then  $(\D^*(W),d)$ forms a complex, whose $p$-th cohomology is denoted by $H^p(\D^*(W))$.
For a closed $p$-current $T$, we denote by $[T]$ its cohomology class.

A form $\o$ in $A^p(W)$ may be naturally thought of as a $p$-current $T_\o$ by
\[
\langle T_\o,\varphi\rangle=\int_W\o\wedge\varphi,\qquad\varphi\in A_c^{m'-p}(W).
\]
If $\o$ is closed, then  $T_\o$ is closed and the assignment $\o\mapsto T_\o$ induces an \iso
\begin{equationth}\label{dRcurr}
\b:H^p_{\rm dR}(W)\stackrel\sim\lra H^p(\D^*(W)).
\end{equationth}

A \v Cech-de~Rham cochain $\sigma$ on a covering $\W$ of $W$ may be also thought of as a current $T_\sigma$ via integration given as \eqref{intCdR}. If $D\sigma=0$, then  $T_\sigma$ is closed and the assignment $\sigma\mapsto T_\sigma$ induces
the \iso\ $\b\circ\a^{-1}:H^p_D(\W)\stackrel\sim\ra H^p(\D^*(W))$.

Also an $(m'-p)$-chain $C$ may be thought of as a $p$-current
$T_C$ by
\[
\langle T_C,\varphi\rangle=\int_C\varphi,\qquad\varphi\in A_c^{m'-p}(W).
\]
If $C$ is a cycle, $T_C$ is closed and the assignment $C\mapsto T_C$ induces an \iso
\[
\gamma:\breve H_{m-p}(W)\stackrel\sim\lra H^p(\D^*(W)).
\]
By \eqref{corrpoin}, we have $\gamma\circ P=\b$. Thus for any closed current $T$ on $W$, there exist
a closed $p$-form $\o$, a \v CdR cocycle $\sigma$ and an $(m'-p)$-cycle $C$ \st

\[
[T]=[T_\o]=[T_\sigma]=[T_C].

\]
We call $\o$, $\sigma$ and $C$, \r,  de~Rham, \v CdR and cycle representatives of $T$.

If $U$ is an open set of $W$, there is a natural inclusion $A^*_c(U)\hra A^*_c(W)$, given by extension by zero, so that we may consider the restriction
$T|_U$ to $U$ of a current $T$ on $W$. The {\em support} $\op{supp}(T)$ of $T$ is the smallest closed subset of $W$ \st\
$T|_{W\ssm \op{supp}(T)}=0$.
\vv

Now we consider the localization problem of currents.
Thus let   $\tilde S$ be  a closed set in $W$. Let $W_0=W\ssm \tilde S$ and  $W_1$  a
\nbd\ of $\tilde S$ in $W$ and consider the covering
$\W=\{W_0, W_1\}$ of $W$.
We have the commutative diagram with exact row (cf. \eqref{longexact})\,:

\begin{equationth}\label{funddiag}
\SelectTips{cm}{}
\xymatrix{H^p_D(\W, W_0)\ar[r]^-{j^*} &H^p_D(\W)\ar[r]^-{\iota^*}\ar[d]^{\b\circ\a^{-1}}_{\wr} &H^p_{\rm dR}(W_0)\ar[d]^{\b}_{\wr}\\
 {} & H^p(\D^{*} (W))\ar[r]^{\iota^*}&H^p(\D^{*} (W_0)).}
\end{equationth}

Suppose $T$ is a closed $p$-current on $W$ \st\ $\iota^*[T]=0$, i.e., $[T|_{W_0}]=0$. Then there is a class $\vP_T$ in $H^p_D(\W, W_0)$ \st\
$[T]=j^\ast\vP_T$. We then say that $T$ is {\em localized at $\tilde S$} and  call $\vP_T$ a {\em  Thom class of $T$ along $\tilde S$}. Here some comments are in order\,:
\begin{enumerate}
\item[(1)]  Any closed current $T$ is localized at $\op{supp}(T)$ in the above sense. It is also localized
at the support of a cycle representive of  $T$.
Thus the set $\tilde S$ as above
may be different from $\op{supp}(T)$,
see Example \ref{smoothcase} below.
\item[(2)]  The class $\vP_T$ is not uniquely determined, as $j^*$ is not injective in general, however in some cases, there is a natural choice of $\vP_T$, see Examples \ref{Thom-submnf} and \ref{Cherncurrent} below.
\end{enumerate}

In the above situation, suppose $\tilde S$ is a compact set satisfying (*).
Let $(0,\psi_1,\psi_{01})$ be a \v CdR representative of $\vP_T$ and $\{\tilde R_0,\tilde R_1\}$ a system of honeycomb cells adapted to $\W$. Then, from the commutativity of the
diagram obtained by replacing $M$ and $\U$ by $W$ and $\W$ in \eqref{PoinAlex} (see also Remark \ref{remimp}), for any closed  form $\varphi$ in $A^{m'-p}_c(W)$ we have\,:
\begin{equationth}\label{localization}
\langle T, \varphi \rangle
=\int_{\tilde R_1}\psi_1\wedge\varphi +\int_{\tilde R_{01}} \psi_{01}\wedge\varphi.
\end{equationth}
Thus the value of $T$ is ``concentrated" near $\tilde S$ and  is explicitly given by the above. Also note that the right
hand side does not depend on the choice of $\vP_T$.

\begin{example}\label{Thom-submnf}
Let $C$ be a finite $(m'-p)$-cycle  in  $W$. Then $T_C$ is localized at
$\tilde S=|C|$ and $\vP_C$ is a natural choice for $\vP_{T_C}$.  In particular, if $C=M$ is a compact oriented submanifold of codimension $p$, then the Thom class $\vP_M$ of $M$ is a
natural choice for $\vP_{T_M}$.
\end{example}

We show that, starting from a closed $p$-form $\o$ \st\ $[T_\o]=[T]$, there is a natural way of constructing
such a class. Thus let $\tilde S$ be a closed set of $W$ and let $\W=\{W_0,W_1\}$ be as before.

\begin{proposition}\label{m1}
Let $T$ be a closed $p$-current on $W$ such that $[T|_{W_0}]=0$ and $\o$ a de~Rham representative of $T$.
Then there exists a Thom class $\vP_T$
which is represented by a \v CdR cocycle of the form $(0,\o,-\psi)$ with $\psi$  a $(p-1)$-form on
$W_{01}$ satisfying $\o=-d\psi$ on
$W_{01}$.
\end{proposition}

\begin{proof} Just to make sure we denote the restrictions of forms explicitly.
From the assumption, there exists a $(p-1)$-form $\psi$ on $W_0$
such that $d\psi=-\o|_{W_0}$. Hence  the cocycle $(\o|_{W_0}, \o|_{W_1}, 0)$  is cohomologous as a \v CdR cocycle to
$(0, \o|_{W_1}, -\psi|_{W_{01}})$ since
\[
(0, \o|_{W_1}, -\psi|_{W_{01}})-(\o|_{W_0}, \o|_{W_1}, 0)=(d\psi, 0, -\psi|_{W_{01}})=D(\psi, 0, 0).
\]
Thus the class $\vP_{T}=[(0, \o|_{W_1}, -\psi|_{W_{01}})]$ satisfies $[T]=j^*\vP_T$.
\end{proof}

In this case, if $\tilde S$ is a compact set satisfying (*), \eqref{localization} is written as
\[
\int_W\o\wedge\varphi=\int_{\tilde R_1} \o\wedge\varphi -\int_{\tilde R_{01}} \psi\wedge\varphi.
\]
Thus the value of the integral away from $\tilde R_1$ is cut off  and is compensated by an integral on $\tilde R_{01}$.

\begin{example}\label{smoothcase} Let $C$ be an $(m'-p)$-cycle  in  $W$ and
 $\o$  a de~Rham representative of $C$.
Let $\tilde S$ be the support of $C$ and set $W_0=W\ssm \tilde S$. Then $T_\o$ is localized at
$\tilde S$ as $[T_\o]=[T_C]$, although we do not have any precise information
about $\op{supp}(T_{\o})$.
Its   Thom class  $\vP_{T_{\o}}$
along $\tilde S$ is represented  by a \v CdR cocycle of the form
$(0, \o, -\psi)$.
\end{example}

\begin{remark} Let $C$, $\o$ and  $\tilde S$ be as in Example \ref{smoothcase}.
Then there is a $(p-1)$-current $R$ \st
\[
T_C-T_\o=dR.
\]
We may think of $R$ as the current defined by a
$(p-1)$-form $\psi$ on $W\ssm \tilde S$ that can be extended as a locally integrable $L^1$ form on $W$
and with $d\psi=-\o$ on $W\ssm \tilde S$. The equation above becomes then
\[
dT_\psi-T_{d\psi}=T_C,
\]
which is a residue formula (cf. \cite[Ch.3,1]{GH}), and the identitiy
\[
D(\psi, 0, 0)+(\o, \o, 0)=(0, \o, -\psi)
\]
may be thought of as the corresponding expression in terms of \v CdR cochains.
\end{remark}

\begin{example}\label{Cherncurrent}
Let $\pi:E\ra W$ be a $C^\infty$ complex vector bundle of rank $r$ and $\na$ a connection for $E$. For $q=0,\dots,r$,
we have the $q$-th Chern form $c_q(\na)$, which is a closed $2q$-form defining the $q$-th Chern class $c_q(E)$
 in
$H^{2q}_{\rm dR}(W)$. We call $T_{c_q(\na)}$ the $q$-th {\em Chern current associated with $\na$}.
Suppose  $E$ admits  $\ell$ sections $\bm{s}=(s_1,\ldots,s_\ell)$
that are linearly independent on the complement of a closed set $\tilde S\subset W$. Then we see that $T_{c_q(\na)}$
is localized at $\tilde S$ and there is a natural
way of choosing a Thom class along $\tilde S$ for $q=r-\ell+1,\dots,r$.

For this, we take an $\bm{s}$-trivial connection $\na_0$ for $E$ on $W_0=W\ssm \tilde S$, i.e., a connection satisfying
$\na_0 s_i=0$ for $i=1,\ldots, \ell$.
Denoting by $c_q(\na_0, \na)$ the  Bott difference form (cf. \cite{Bo1}, \cite{Su6}), we have
$c_q(\na)|_{W_0}-c_q(\na_0)=d\, c_q(\na_0, \na)$.
Since the connection $\na_0$ is
$\bm{s}$-trivial, it follows that $c_q(\na_0)=0$ so that $c_q(\na)|_{W_0}$ is exact for $q=r-\ell+1,\ldots, r$.
Hence the Chern current $T_{c_q(\na)}$ localizes at $\tilde S$.
As its Thom class along $\tilde S$, we may take the class  $c_q(E,\bm{s})$ in
$H^{2q}(W,W\ssm \tilde S)\simeq H^{2q}_D(\W,W_0)$ represented by the cocycle (cf. Proposition \ref{m1})\,:
\[
(0,c_q(\na)|_{W_1}, c_q(\na_0, \na)).
\]
 This class does not depend on the choice of $\na$
or $\na_0$ (cf. \cite[Ch.III, Lemma 3.1]{Su6}) and is a natural choice of Thom class for  $T_{c_q(\na)}$.
It is the {\em localization of $c_q(E)$ at $\tilde S$ by $\bm{s}$}.

The Thom class of a complex \vb\ as a real oriented bundle may be expressed in this manner
(cf. \cite[Ch.III, Theorem 4.4]{Su6}).
\end{example}

\subsection{Localized intersection of currents} \label{sec:localized}

Let $W$ be an oriented  $C^\infty$ manifold of dimension $m'$ as before.
For  closed currents $T_1$ and  $T_2$,
it is possible to define the intersection product $T_1\cdot T_2$ in the homology of $W$ using the \iso\ $\b$
and the Poincar\'e duality (cf. \eqref{dRcurr}, \eqref{defint}). Also if $T_1$ and $T_2$ are localized at compact sets
$\tilde S_1$ and $\tilde S_2$ satisfying (*),
we may define the localized intersection $(T_1\cdot T_2)_{\tilde S}$ in the homology of $\tilde S=\tilde S_1\cap \tilde S_2$ using Thom
classes $\vP_{T_1}$ and $\vP_{T_2}$ and the Alexander duality (cf. \eqref{locint}).

Here we consider the case $T_1=T_M$  with $M$ a compact oriented sub\mfd\ of
dimension $m$ in $W$ and  obtain a residue theorem on $M$.
In the sequel we take the Thom class $\vP_M$ of $M$ (cf. Subsection \ref{Thom} (b))   as $\vP_{T_M}$ and set $k=m'-m$. Recall that by the Alexander \iso
\[
A_{W,M}:H^k(W,W\ssm M)\stackrel\sim\lra  H_m(M),
\]
the class $\vP_M$ corresponds to the fundamental class $M$. Let $i: M \hra W$ denote the  inclusion.

\paragraph{First localization\,:} Let $c$ be a class in $\breve H_{m'-p}(W)$. Recall that we have the intersection product
$(M\cdot c)_M$ localized at $M$ (cf. \eqref{locint}), which is a class in $H_{m-p}(M)$ defined as
$A_{W,M}(\vP_M\smallsmile P_W^{-1}c)$. We denote it by $M\cdot c$\,:
\[
M\cdot c:=(M\cdot c)_M.
\]
It is sent to $[M]\cdot c$ by $i_*: H_{m-p}(M)\ra\breve H_{m-p}(W)$.

\paragraph{Second localization\,:} Let $\tilde S$ be a compact set  of $W$ satisfying (*).  We have the Alexander
\iso
\[
A_{W,\tilde S}:H^p(W,W\ssm \tilde S)\stackrel\sim\lra  H_{m'-p}(\tilde S).
\]
We set $S=\tilde S\cap M$ and suppose it also satisfies (*).
For a class $c$ in $H_{m'-p}(\tilde S)$, we have the
class $(M\cdot c)_S$ in $H_{m-p} (S)$ (cf. \eqref{locint}).

\begin{proposition}\label{diagfund} The following diagrams are  commutative\,:
\[
\SelectTips{cm}{}
\xymatrix{ H^p(W) \ar[d]_{i^*}\ar[r]^-\sim_-{P_W}&\breve H_{m'-p}(W)\ar[d]^{M\cdot\ }\\
 H^p(M)\ar[r]^-\sim_-{P_M}&   H_{m-p}(M),}\hspace{1.7cm}
 \xymatrix{ H^p(W, W\ssm \tilde S) \ar[d]_{i^*}\ar[r]^-\sim_-{A_{W,\tilde S}}&\breve H_{m'-p}(\tilde S)\ar[d]^{(M\cdot\ )_S}\\
 H^p(M, M \ssm S)\ar[r]^-\sim_-{A_{M,S}}&   H_{m-p}(S).}
\]
\end{proposition}

\begin{proof} We prove the commutativity of the second diagram,  the proof for the first one  being  similar.
We  have the cup product followed by the Alexander isomorphism\,:
\[
H^k(W, W\ssm  M)\times H^p(W, W\ssm  \tilde S) \stackrel\smallsmile\lra
  H^{k+p}(W, W\ssm  S)\stackrel{A_{W, S}}\lra H_{m-p}(S).
\]
Nothing that the Alexander \iso\ is given by the left cap product with the fundamental class and using properties of cap and cup products, we have,
for a class $u$ in $H^p(W, W\ssm  \tilde S)$,
\[
  A_{M, S}(i^*u) =A_{W, S}(\vP_M\smallsmile u)=(M\cdot A_{W,\tilde S}u)_S.
\]
\end{proof}

In view of the above,  we define intersection products in a more general situation where $M$ is not necessarily a sub\mfd\ of $W$\,:

\begin{definition}\label{prodmap} Let $W$ and $M$ be oriented $C^\infty$ \mfd s of dimensions $m'$ and $m$, \r, and $F:M\ra W$  a $C^\infty$ map.  We define the intersection product $M\cdot_F\ $ so that the first diagram
below is commutative. Also, for a compact set $\tilde S$ satisfying (*) in $W$, we set $S=F^{-1}(\tilde S)$ and suppose
$S$ is compact and satisfy (*).  We then define the localized intersection product $(M\cdot_F\ )_S$
so that the second diagram is commutative\,:
\[
\SelectTips{cm}{}
\xymatrix{ H^p(W) \ar[d]_{F^*}\ar[r]^-\sim_-{P_W}&\breve H_{m'-p}(W)\ar[d]^{M\cdot_F\ }\\
 H^p(M)\ar[r]^-\sim_-{P_M}& \breve  H_{m-p}(M),}\hspace{1.7cm}
 \xymatrix{ H^p(W, W\ssm \tilde S) \ar[d]_{F^*}\ar[r]^-\sim_-{A_{W,\tilde S}}&\breve H_{m'-p}(\tilde S)\ar[d]^{(M\cdot_F\ )_S}\\
 H^p(M, M \ssm S)\ar[r]^-\sim_-{A_{M,S}}&   H_{m-p}(S).}
\]
\end{definition}

\begin{remark} 1.  Let $M$ be  a  sub\mfd\ of $W$ and $i:M\hra W$  the inclusion.
If $M$ is compact,  $M\cdot_i \ $ is the product
 $M\cdot\ $ defined before. We may also define the  product $M\cdot\ $ as $M\cdot_i \ $ in the case
 $M$ is not compact.
\smallskip

\noindent
2. The products as above are defined in the algebraic category in \cite{Fu}.
\end{remark}

For a closed $p$-current $T$ on $W$, we define
\[
M\cdot_FT:=M\cdot_FP\b^{-1}[T].
\]
Suppose $T$ is localized at $\tilde S$. Then taking a Thom class $\vP_{T}$   of $T$ along $\tilde S$, we define
the {\em residue}  of $\vP_{T}$ on $M$ at $S$ by
\[
\op{Res}(F^*\vP_T,S):=(M\cdot_FA(\vP_T))_S.
\]

Suppose $S$ has a finite number of connected components $(S_\l)_\l$. Then we have a decomposition
$H_{m-p} (S)=\bigoplus_\l H_{m-p} (S_\l)$ and accordingly $\op{Res}(F^*\vP_T,S)$ determines
a class in $H_{m-p} (S_\l)$, which  is denoted by $\op{Res}(F^*\vP_T,S_\l)$.
We can state the following  general residue theorem, which follows from the commutativity of the diagram \eqref{PoinAlex}\,:

\begin{theorem}\label{res-m} Let $W$ and $M$ be oriented $C^\infty$ \mfd s of dimensions $m'$ and $m$, \r, and $F:M\ra W$  a
$C^\infty$ map.  Let $T$ be a closed $p$-current on $W$ \st\ $[T|_{W\ssm \tilde S}]=0$ for some  compact subset
$\tilde S$ satisfying (*) in $W$.
Suppose that  $S=F^{-1}\tilde S$ is compact, satisfies (*) and  has a finite number of connected components $(S_\l)_\l$. Then

\begin{enumerate}
\item[{\rm (1)}] For each $\l$ we have a class $\op{Res}(F^*\vP_T,\, S_\l)$ in $H_{m-p}(S_\l)$.
\item[{\rm (2)}] We have the ``residue formula"\,:
\[
M\cdot_F T=\sum_\l (i_\l)_*\op{Res}(F^*\vP_T,\, S_\l)\qquad\text{in}\ \ \breve H_{m-p}(M),
\]
where $i_\l:S_\l\hra M$ denotes the inclusion.
\end{enumerate}
\end{theorem}

We may express $M\cdot_F T$ and $\op{Res}(F^*\vP_T,\, S_\l)$ as follows.
Let $T$ be a closed $p$-current on $W$ and $\o$ a de~Rham representative of $T$. From \eqref{corrpoinform}
and \eqref{corrpoin}, we have\,:

\begin{proposition}\label{prodindR} {\rm (1)} The intersection product $M\cdot_F T$ in $\breve H_{m-p}(M)$ is represented
by a cycle $C$ \st\
\[
\int_MF^*\o\wedge\varphi=\int_C\varphi
\]
for any closed form $\varphi$ in $A_c^{m-p}(M)$.
\vv

\noindent
{\rm (2)} In the \iso\
$\breve H_{m-p}(M)\simeq H^{m-p}_c(M)^*$,  $M\cdot_F T$ corresponds to the functional on $H^p_c(M)$ that
assigns to $[\varphi]$ the left hand side above.
\vv

\noindent
{\rm (3)}
In particular, if $p=m$ and if $M$ is compact, $M\cdot_F T$ is a number given by
\[
M\cdot_F T=\int_MF^*\o.
\]
\end{proposition}

Suppose $T$ satisfies the conditions in Theorem \ref{res-m}. Let $W_0=W\ssm \tilde S$ and $W_1$ a \nbd\ of $\tilde S$ and
consider the covering $\W=\{W_0,W_1\}$.
Let $\vP_T$ be represented by a \v CdR cocycle $(0,\psi_1,\psi_{01})$ in $A^p(\W,W_0)$.
For each $\l$ we take a regular \nbd\ $U_\l$ of $S_\l$ in $M$ \st\
$F(U_\l)\subset W_1$ and that $U_\l\cap U_\mu=\emptyset$ if $\l\ne\mu$.
For each $\l$, we take a compact sub\mfd\ $R_\l$ of dimension $m$ with $C^\infty$ boundary in $U_\l$, containing $S_\l$ in its interior.
From \eqref{corralexform} and \eqref{corralex}, we have\,:

\begin{proposition}\label{propint} {\rm (1)} The residue $\op{Res}(F^*\vP_T,\, S_\l)$ in $H_{m-p}(S_\l)$ is represented
by a cycle $C$ \st\
\[
\int_{R_\l}F^*\psi_1\wedge\varphi+\int_{R_{0\l}}F^*\psi_{01}\wedge\varphi=\int_C\varphi
\]
for any closed form $\varphi$ in $A^{m-p}(U_\l)$.
\vv

\noindent
{\rm (2)} In the \iso\
$H_{m-p}(S_\l)\simeq H^{m-p}_{\rm dR}(U_\l)^*$,  $\op{Res}(F^*\vP_T,\, S_\l)$ corresponds to the
 functional on $H^{m-p}_{\rm dR}(U_\l)$ that assigns to $[\varphi]$ the left hand side above.
\vv

\noindent
{\rm (3)} In particular, if $p=m$, the residue is a number given by
\[
\op{Res}(F^*\vP_T,\, S_\l)=\int_{R_\l}F^*\psi_1+\int_{R_{0\l}}F^*\psi_{01}.
\]
\end{proposition}

\begin{example}\label{rescycle}
Let $C$ be a finite $(m'-p)$-cycle on $W$, $\tilde S=|C|$ and $S=F^{-1}\tilde S$. We take $\vP_C$ as $\vP_{T_C}$. Then
\[
M\cdot_F T_C=M\cdot_F [C],\qquad \op{Res}(F^*\vP_C,\, S_\l)=(M\cdot_F C)_{S_\l}
\]
and the residue formula becomes
\[
M\cdot_F [C]=\sum_\l(i_\l)_*(M\cdot_F C)_{S_\l}\qquad\text{in}\ \ \breve H_{m-p}(M).
\]
In particular, if $M$ is compact and $p=m$,
\[
M\cdot_F [C]=\sum_\l(M\cdot_F C)_{S_\l}.
\]

Let $\o$ be a de~Rham representative of $C$. Then $T_\o$ is localized at $\tilde S$. As $\vP_{T_\o}$ we may take
the class represented by a cocycle of the form $(0,\o,-\psi)$ (cf. Example \ref{smoothcase}). As a homology class,
$M\cdot_F T_\o=M\cdot_F [C]$. As a functional, it is given as in Proposition \ref{prodindR}. Also $\op{Res}(F^*\vP_{T_\o},\, S_\l)$
is a functional given as in Proposition \ref{propint} with $\psi_1=\o$ and $\psi_{01}=-\psi$.
\end{example}

See Propositions \ref{propisolated} and \ref{coinnonisol} below for  explicit expressions of $(M\cdot_F C)_{S_\l}$ in some special cases.

\begin{example}\label{exmult}
Let $W$ be a complex \mfd\ of dimension $n'$ and $M$ a complex sub\mfd\ of dimension $n$.
Also let $V$ be an analytic subvariety of $W$ of dimension $k$. Recall that there exists a subanalytic triangulation of $W$
compatible with $M$, $V$ and $\op{Sing}(V)$, the singular set of $V$. Thus $V$ may be thought of as a chain, which is not $C^\infty$
but still has the associated current $T_V$ of integration.
Moreover it is a cycle, as the real codimension of $\op{Sing}(V)$ in $V$ is greater than or equal to two. If $n+k=n'$ and if $p$ is an
isolated point of $M\cap V$, we have
\[
(M\cdot V)_p\ge \op{mult}_{p}(V),
\]
the multiplicity of $V$ at $p$. The equality holds, by definition, if $M$ is general with respect to $V$, i.e., the intersection of the
tangent space of $M$ at $p$
and the tangent cone of $V$ at $p$ consists only of $p$.
Note that $\op{mult}_{p}(V)$ coincides with the Lelong number of $T_V$ at $p$ (e.g., \cite[Ch.3, 2]{GH}).
\end{example}

We finish this section by giving a formula for the residue at  a non-isolated component. Thus,  in the  situation of
Theorem \ref{res-m}, suppose that $S_\l$ is an oriented sub\mfd\ of $M$
of dimension $m-p$ with orientation compatible with that of $M$ in the sense described in Subsection \ref{Thom} (b). Let $p_\l$ be a point in $S_\l$ and $B_\l$ a small open ball of dimension $p$ in $M$ transverse
to $S_\l$ at $p_\l$. We orient $B_\l$ so that the orientation of $B_\l$ followed by that of $S_\l$ gives the
orientation of $M$. Setting $F_\l=F|_{B_\l}$, we have the commutative diagram
\[
\SelectTips{cm}{}
\xymatrix{ H^p(W,W\ssm \tilde S) \ar[d]_{F_\l^*}\ar[r]^-\sim_-{A_{W,\tilde S}}&H_{m'-p}(\tilde S)
\ar[d]^{(B_\l\cdot_{F_\l}\ )_{p_\l}} \\
 H^p(B_\l,B_\l\ssm p_\l)\ar[r]^-\sim_-{A_{B_\l,p_\l}}& H_0 (p_\l).}
\]
We have the residue $\op{Res}(F_\l^*\vP_T,p_\l)=(B_\l\cdot_{F_\l} A(\vP_T))_{p_\l}$ in $H_0 (p_\l)\simeq\C$ so that it is a number.

\begin{theorem}\label{resnoniso} In the  situation of Theorem \ref{res-m}, suppose that $S_\l$ is an oriented  sub\mfd\ of $M$ of dimension $m-p$ and let $p_\l$  and $F_\l$ be as above. Then we have\,:
\[
\op{Res}(F^*\vP_T,S_\l)=\op{Res}(F_\l^*\vP_T,p_\l)\cdot S_\l\qquad\text{in}\ \ H_{m-p}(S_\l).
\]
\end{theorem}

\begin{proof} We try to find $\op{Res}(F^*\vP_T,S_\l)$ by  Proposition \ref{propint}. As $U_\l$, we take a tubular \nbd\ of $S_\l$ with a $C^\infty$ projection
$\pi:U_\l\ra S_\l$, which gives $U_\l$ the structure of a bundle of open balls of dimension $p$. Setting $U_0=U_\l\ssm S_\l$, we consider the covering $\U_\l=\{U_0,U_\l\}$ of  $U_\l$.
As $R_\l$, we take a bundle on $S_\l$ of closed balls of dimension $p$ in $U_\l$.
Then $R_{0\l}$ is a bundle on $S_\l$ of $(p-1)$-spheres.
We denote the restrictions of $\pi$ to $R_\l$ and $R_{0\l}$ by $\pi_\l$ and $\pi_{0\l}$, \r.
For a closed $(m-p)$-form $\varphi$  on $U_\l$, we compute the integral
\[
I:=\int_{R_\l}F^*\psi_1\wedge\varphi+\int_{R_{0\l}}F^*\psi_{01}\wedge\varphi.
\]

Since $\pi$ induces an \iso\ $\pi^*:H^{m-p}_{\rm dR}(S_\l)\stackrel\sim\ra H^{m-p}_{\rm dR}(U_\l)$, there
exist a closed $(m-p)$-form $\t$ on $S_\l$ and an $(m-p-1)$-form $\tau$ on $U_\l$ \st\
\[
\varphi=\pi^*\t+d\tau.
\]
Using the projection formula, the fact that $dF^*\psi_1=0$ and the Stokes formula, we have
\[
\int_{R_\l}F^*\psi_1\wedge\varphi=\int_{S_\l}(\pi_\l)_*F^*\psi_1\cdot\t+
(-1)^{p+1}\int_{R_{0\l}}F^*\psi_1\wedge\tau,
\]
where $(\pi_\l)_*$ denotes the integration along the fiber of $\pi_\l$. Note that $(\pi_\l)_*F^*\psi_1$ is a
$C^\infty$ \fcn\ on $S_\l$. Noting that $dF^*\psi_{01}=F^*\psi_1$ on $U_{0\l}$ and $\partial R_{0\l}=\emptyset$, we also compute to get
\[
\int_{R_{0\l}}F^*\psi_{01}\wedge\varphi=\int_{S_\l}(\pi_{0\l})_*F^*\psi_{01}\cdot\t+
(-1)^p\int_{R_{0\l}}F^*\psi_1\wedge\tau.
\]
Thus we have
\[
I=\int_{S_\l}((\pi_\l)_*F^*\psi_1+(\pi_{0\l})_*F^*\psi_{01})\cdot\t
\]

Now recall that we have the integration along the fiber on the \v CdR cochains\,:
\[
\pi_*:A^q(\U_\l,U_0)\lra A^{q-p}(S_\l),
\]
which assigns to $\sigma=(0,\sigma_\l,\sigma_{0\l})$ the form $(\pi_\l)_*\sigma_\l+(\pi_{0\l})_*\sigma_{0\l}$
on $S_\l$. Moreover it is compatible with the differentials $D$ and $d$ (cf. \cite[Ch.II, 5]{Su6}). Since
 $(0, F^*\psi_1|_{U_\l},F^*\psi_{01}|_{U_{0\l}})$ is a \v CdR cocycle
in $A^p(\U_\l,U_0)$, the \fcn\ $(\pi_\l)_*F^*\psi_1+(\pi_{0\l})_*F^*\psi_{01}$ is $d$-closed so that it is a
constant. By definition the constant is exactly $\op{Res}(F_\l^*\vP_T,p_\l)$ above. Finally from
\[
\int_{S_\l}\t=\int_{S_\l}\varphi,
\]
we have the theorem.
\end{proof}

\begin{remark} 1. In the above situation,  $H_{m-p}(S_\l,\Z)\simeq\Z$ and is generated by the fundamental class $S_\l$. Thus if
$\op{Res}(F_\l^*\vP_T,p_\l)$ is an integer, $\op{Res}(F^*\vP_T,S_\l)$ is an integral class.
\smallskip

\noindent
2. The above theorem  can also be proved topologically as \cite[Theorem 4.1.1, see also Theorem 7.3.2]{Su8}.
In fact, using  techniques and results in \cite{Su8}, we may compute residues in various settings.
\end{remark}

\section{Coincidence point theorem}\label{seccoin}

\subsection{Coincidence homology classes and   indices}\label{local-coin}

Let $M$ and $N$ be connected and oriented  $C^\infty$ \mfd s of dimensions $m$ and $n$, \r, with $m\ge n$.
We set $W=M\times N$ and
orient $W$ so that the orientation of $M$ followed by that of $N$ gives the orientation of $W$.
Let $f,\, g:M\ra N$ be $C^\infty$ maps
and denote by $\vG_f$ and $\vG_g$ the graphs of $f$ and  $g$  in $W$. We consider the map
\[
\tilde f:M\lra\vG_f\subset W\qquad\text{defined by}\ \ \tilde f(x)=(x,f(x)),
\]
which is a diffeomorphism onto $\vG_f$. We orient $\vG_f$ so that $\tilde f$  is orientation preserving.
Similarly we define $\tilde g$ for $g$. Recall the diagram (cf. Definition \ref{prodmap})\,:
\[
\SelectTips{cm}{}
\xymatrix{ H^n(W)  \ar[d]_{\tilde f^*}\ar[r]^-\sim_-{P}&\breve H_m(W)
\ar[d]^{M\cdot_{\tilde f}\ } \\
 H^n(M)\ar[r]^-\sim_-{P}&  \breve H_{m-n} (M).}
\]

\begin{definition}\label{gcoin} The {\em global coincidence class $\vL(f,g)$  of the pair $(f,g)$} is defined by
\[
\vL(f,g)=M\cdot_{\tilde f}[ \vG_g]\qquad\text{in}\ \ \breve H_{m-n}(M).
\]
\end{definition}

Note that $\tilde f$ induces an \iso\ $\tilde f_*:\breve H_{m-n}(M)\stackrel\sim\ra \breve H_{m-n}(\vG_f)$
and   $\vL(f,g)$ corresponds to $\vG_f\cdot [ \vG_g]$ in $\breve H_{m-n}(\vG_f)$, which is sent to
$[\vG_f]\cdot [ \vG_g]$ in $\breve H_{m-n}(W)$ by the canonical \homo\
$\breve H_{m-n}(\vG_f)\ra\breve H_{m-n}(W)$.

We define the {\em coincidence point set} of the pair $(f,g)$ by
\[
\op{Coin}(f,g)=\{ \,p \in M\mid f(p)=g(p) \,\}.
\]
Note that $\op{Coin}(f,g)=\tilde f^{-1}(\vG_g)$. For shortness, we set $S=\op{Coin}(f,g)$.

From now on we assume that $M$ is compact
so that $\vG_g$ and $S$ are compact. Recall the diagram (cf. Definition \ref{prodmap})\,:
\[
\SelectTips{cm}{}
\xymatrix{ H^n(W,W\ssm \vG_g) \ar[d]_{\tilde f^*}\ar[r]^-\sim_-{A}&  H_m(\vG_g)
\ar[d]^{(M\cdot_{\tilde f}\ )_S} \\
H^n(M,M\ssm S)\ar[r]^-\sim_-{A}&  H_{m-n} (S).}
\]

\begin{definition}\label{lcoin}
The {\em local coincidence class $\vL(f,g;S)$ of the pair $(f,g)$ at $S$} is defined to be the localized intersection class\,:
\[
\vL (f,g;S)=(M\cdot_{\tilde f} \vG_g)_S\qquad\text{in}\ \  H_{m-n}(S).
\]
\end{definition}

Note that $\tilde f$ induces a \homo\ $\tilde f_*:H_{m-n}(S)\ra H_{m-n}(\vG_f)$
and   $\vL(f,g;S)$  is sent to
$\vG_f\cdot [ \vG_g]$.

\begin{remark} 1. The classes $\vL(f,g)$ and $\vL(f,g;S)$ are in fact in homology with $\Z$ coefficients.
\smallskip

\noindent
2. We have
\[
\vL(g,f)=(-1)^n\vL(f,g),\qquad \vL(g,f;S)=(-1)^n\vL(f,g;S).
\]
\end{remark}

Suppose $S=\op{Coin}(f,g)$ has a finite number of connected components $(S_\l)_\l$.
Then we have $H_{m-n}(S)=\bigoplus_\l H_{m-n}(S_\l)$ and accordingly we have the
local coincidence class $\vL(f,g; S_\l)$ in $H_{m-n}(S_\l)$. From Theorem \ref{res-m}, we have
a general  coincidence point theorem\,:

\begin{theorem}\label{gcth} In the above situation
\[
\vL(f,g)=\sum_\l (i_\l)_*\vL(f,g; S_\l)\qquad\text{in}\ \  H_{m-n}(M).
\]
\end{theorem}

In general, $\vL(f,g)$ and $\vL(f,g; S_\l)$ are given as in Propositions \ref{prodindR} and \ref{propint}.
The theorem becomes more meaningful if we have explicit descriptions of them.

In the case  $m=n$, $\vL(f,g; S_\l)$ is in $H_0(S_\l)=\C$ so that it is a number
(in fact an integer), which we call the {\em coincidence index} of $(f,g)$ at $S_\l$. If $S_\l$ consists of a point $p$, we have the following explicit formula.
In fact it is already known, however we give an alternative short  proof using the  Thom class in the \v Cech-de~Rham cohomology.
Let $U$ be a coordinate \nbd\ around $p$ with
coordinates $x=(x_1,\dots,x_m)$ in $M$ and $V$ a coordinate \nbd\ around $f(p)=g(p)$ with
coordinates $y=(y_1,\dots,y_m)$ in $N$. Also let $D$ be a closed ball around $p$ in $U$ \st\ $f(D)\subset V$ and $g(D)\subset V$.
Thus we may consider the map $g-f:D\ra\R^m$ whose image is the origin $0$ in $\R^m$ only at $p$. The boundary $\partial D$
is homeomorphic to the unit sphere $S^{m-1}$ and we have the map
\[
\gamma:\partial D\lra S^{m-1}\qquad\text{defined by}\ \ \gamma(x)=\frac{g(x)-f(x)}{\Vert g(x)-f(x)\Vert}.
\]
We denote the degree of this map by $\op{deg}(g-f,p)$.

\begin{proposition}\label{propisolated} In the above situation
\[
\vL(f,g; p)=\op{deg}(g-f,p).
\]
\end{proposition}

\begin{proof} Let $\vP_{\vG_g}$ be the Thom class of $\vG_g$ and $(0,\psi_1,\psi_{01})$ its \v CdR representative.
We may take $D$ as $R_\l$ in Proposition \ref{propint}. Since $R_{0\l}=-\partial D$, we have
\begin{equationth}\label{coiniso}
\vL(f,g; p)=\int_D\tilde f^*\psi_1-\int_{\partial D}\tilde f^*\psi_{01}.
\end{equationth}
Recall that $\vP_{\vG_g}$ may be naturally identified with the Thom class of  the normal bundle $N_{\vG_g}$,
which is trivial over $\tilde g(U)$\,;
$N_{\vG_g}|_{\tilde g(U)}\simeq \R^m\times \tilde g(U)$ (cf. Subsection \ref{Thom} (b)). Let
$\rho_g:N_{\vG_g}|_{\tilde g(U)}\ra \R^m$ denote the projection onto the fiber direction.  Also let $\psi_m$ be
an angular form on $\R^m\ssm\{0\}$. Then on $\tilde g(U)$ the Thom class of $N_{\vG_g}$ is represented by the cocycle (cf. Proposition \ref{Thomloc})
\[
(0,0,-\rho_g^*\psi_m).
\]

Let $\pi_1$ and $\pi_2$ denote the projections of $W=M\times N$ onto the first and second factors, respectively. We set
$x=\pi_1^*x$ and $y=\pi_2^*y$ on $U\times V$. By our orientation convention, in a \nbd\ of $\tilde g(p)$ in $N_{\vG_g}$, we may take $g(x)-y$ as fiber coordinates and $x$ as base coordinates of the bundle $N_{\vG_g}$ so that we may write
$\rho_g(x,y)=g(x)-y$. Then \eqref{coiniso} becomes
\[
\vL(f,g; p)=\int_{\partial D}(\rho_g\circ\tilde f)^*\psi_m=\int_{\partial D}(g-f)^*\psi_m,
\]
which is nothing but $\op{deg}(g-f,p)$.
\end{proof}

In the above situation, let $J_f(p)$ and
$J_g(p)$ denote the Jacobian matrices of $f$ and $g$ at $p$. A coincidence point $p$ of the pair $(f,g)$ is said to be {\em non-degenerate} if
\[
\det(J_g(p)-J_f(p))\ne 0.
\]

\begin{corollary} If $p$ is a non-degenerate coincidence point,
\[
\vL(f,g; p)=\op{sgn}\, \det(J_g(p)-J_f(p)).
\]
\end{corollary}

Now consider the case $m>n$. Suppose $S_\l$ is an oriented sub\mfd\ of $M$ of dimension $m-n$.  Let $p_\l$ be a point in $S_\l$ and $B_\l$ a small open ball of dimension $n$ in $M$ transverse
to $S_\l$ at $p_\l$. Setting $f_\l=f|_{B_\l}$ and $g_\l=g|_{B_\l}$, we have $\deg(g_\l-f_\l,p_\l)$.
From Theorem \ref{resnoniso}, we have\,:

\begin{proposition}\label{coinnonisol} Let $S_\l$ be a connected component of $\op{Coin}(f,g)$.
If $S_\l$ is an oriented sub\mfd\ of $M$ of dimension $m-n$,
\[
\vL(f,g;S_\l)=\op{deg}(g_\l-f_\l,p_\l)\cdot S_\l\qquad\text{in} \ \  H_{m-n}(S_\l).
\]
\end{proposition}

\subsection{Lefschetz coincidence point formula}

Let $M$ and $N$ be compact, connected and oriented $C^\infty$ manifolds of the same dimension $m$ and
let $f, g \colon M \ra N$ be $C^\infty$ maps. In this situation, $\breve H_0(M)=H_0(M)=\C$
and $\vL(f,g)$ is a number (in fact an integer), which has an explicit description.
Let
\[
H^p(f):H^p(N)\lra H^p(M)
\]
be the \homo\ induced by $f$ on the $p$-th  cohomology group and similarly for $H^p(g)$. We set $q=m-p$. The
Poincar\'e duality  allows us to define the composition
\[
\begin{CD}
H^q(M)\simeq H^p(M)^* @>{H^p(g)^*}>> H^p(N)^* \simeq H^q(N) @>{H^q(f)}>> H^q(M).
\end{CD}
\]
We define the {\em Lefschetz coincidence number $L(f,g)$ of the pair $(f,g)$} as
\[
L(f,g):=\sum_{q=0}^m (-1)^q \cdot \op{tr} (H^q(f) \circ H^{m-q}(g)^*).
\]

Although the following is already known, we include a proof for the sake of completeness. It is  a modification of the presentation as given in \cite{GH}
for the fixed point case, i.e., the case $M=N$ and $g=1_M$, the identity map of $M$.

\begin{proposition}\label{globallef}
In the above situation we have\,:
\[
\vL(f,g)=L(f,g).
\]
\end{proposition}

\begin{proof}
Let $\{\mu_{i}^{p} \}_i$ be a set of closed forms representing a basis of $H_{\rm dR}^p(M)$.
We set $q=m-p$ and let $\{\check{\mu}^{q}_{j} \}_j$ be a set of forms  representing a basis of $H_{\rm dR}^q(M)$ dual
 to $\{[\mu_{i}^{p}]\}_i$\,:
\[
\int_M\mu_{i}^{p}\wedge \check{\mu}_{j}^{q}=\delta_{ij}.
\]
We also take a set of forms  $\{\nu_k^{p} \}_k$ representing a basis of $H_{\rm dR}^p(N)$
and  a set of forms $\{\check{\nu}^{q}_\ell \}_\ell$ representing a  basis of $H_{\rm dR}^q(N)$ dual  to
$\{[\nu_k^{p}] \}_k$.
By the K\"unneth formula, a basis of $H_{\rm dR}^m(W)$, $W=M\times N$,  is represented by
\[
\left\{\,  \xi^{p,q}_{i,\ell}=\pi_1^* \mu_{i}^{p}\wedge \pi_2^* \check\nu_\ell^{q}\, \right\}_{p+q=m},
\]
where $\pi_1$ and $\pi_2$ are projections onto the first and second factors.

Note that in general, for a $p$-form $\o$ on $M$ and a $q$-form $\t$ on $N$, we have
\begin{equationth}\label{intprod}
\int_{\vG_f}\pi_1^*\o\wedge\pi_2^*\t=\int_M\o\wedge f^*\t
\end{equationth}
and similarly for the integration on $\vG_g$.

Let $G^p=(g^p_{ki})$ be the matrix representing $H^p(g)$ in the  bases
$\{[\nu_k^{p}] \}_k$ and $\{[\mu_{i}^{p}] \}_i$\,:
\[
H^p(g)[\nu_k^p]=\sum_i g^p_{ik}[\mu_i^p].
\]
Thus the dual map $H^p(g)^*$  is represented by the transposed ${}^t\hskip-.2mm G^p$ in the  bases
$\{[\check\mu_j^q] \}_j$ and $\{[\check\nu_\ell^{q}] \}_\ell$.
Also let $\check F^q=(\check f^q_{\ell j})$ be the matrix representing
$H^q(f)$  in the  bases  $\{[\check\nu_\ell^{q}] \}_\ell$ and $\{[\check\mu_j^q \}_j$\,:
\[
H^q(f)[\check\nu_\ell^{q}]=\sum_j\check f^q_{j\ell}[\check\mu_j^q].
\]

Let $\eta_g$ be an $m$-form representing the Poincar\'e dual of $[\vG_g]$ in $W$.
Using \eqref{corrpoin} and \eqref{intprod} for $\vG_g$, we compute to get
\[
[\eta_g]=\sum_{q,i,\ell}(-1)^q g^{p}_{i\ell}\, [\xi^{p,q}_{i,\ell}].
\]
Thus we have
\[
\vL(f,g)=\int_M\tilde f^*\eta_g
=\sum_{q,i,\ell}(-1)^{q}g^{p}_{i\ell}\int_M \mu_{i}^{p}\wedge f^*\check\nu_\ell^{q}
=\sum_{q,i,\ell}(-1)^{q}g^{p}_{i\ell}\,  \check f^q_{i\ell}.
\]
Since $ \check f^q_{i\ell}$ is the $\ell i$ entry of $\check F^q$ and $g^{p}_{i\ell}$ the $i\ell$ entry of
${}^t\hskip-.2mm G^{p}$, we have the proposition.
\end{proof}

From Theorem \ref{gcth} and  Propositions \ref{propisolated} and \ref{globallef}, we have\,:

\begin{theorem}\label{non-isolatecoin} Let $M$ and $N$ be compact oriented $C^{\infty}$ manifolds of same dimension
and let $f , g \colon M \ra N$ be $C^\infty$ maps.
Suppose $\op{Coin}(f,g)$ has a finite number of connected components $(S_\l)_\l$. Then
\[
L(f,g)=\sum_\l \vL(f,g;S_\l).
\]
\end{theorem}

In  the case  the set of coincidence points consists only of isolated points, we have\,:
\begin{corollary}[Lefschetz coincidence point formula] Let $M$ and $N$ be compact oriented $C^{\infty}$
manifolds of the same dimension
and let $f , g \colon M \ra N$ be $C^\infty$ maps.
Suppose $\op{Coin}(f,g)$ consists   of a finite number of isolated points. Then
\[
L(f,g)=\sum_{p \in \op{Coin}(f,g) }\deg(g-f,p).
\]
Moreover, if all coincidence points are isolated and non-degenerate then
\[
L(f,g)=\sum_{p\in \op{Coin}(f,g) }\op{sgn}\, \det(J_g(p)-J_f(p)).
\]\label{lefcpf}
\end{corollary}

\begin{remark} 1. The above theory applied to the case $N=M$ and $g=1_M$, the identity map of $M$, gives a general
fixed point theorem for $f$ and the Lefschetz fixed point formula, which is  effective also in the study of periodic
points.
\smallskip

\noindent
2. Let  $f,g:M\ra N$ be  $C^\infty$ maps.  If
$g$ is a diffeomorphism, the coincidence theory for the pair $(f,g)$  is equivalent to
the fixed point theory for the map $g^{-1}\circ f$ of $M$.
\smallskip

\noindent
3. As a textbook dealing with the Lefschetz coincidence theory, we list  \cite{V}.
\smallskip

\noindent
4. In \cite{S}, for a pair of maps  from a topological space to a \mfd, the Lefschetz \homo\
is defined and it is proved that if it is non-trivial, then there is a coincidence point.
\end{remark}

config: arxiv


TITLE: Use a substitution to turn $\cos(y)y'+ \tan(x)\sin(y)=\sin(x)$ into a linear differential equation.
QUESTION [1 upvotes]: $$\cos(y)\frac{dy}{dx} + \tan(x)\sin(y)=\sin(x)$$
Is there any method to find a substitution to fit this purpose? Or is it purely done through brute force/trial and error?

REPLY [1 votes]: Hint:
$$\int{\cos(y)\frac{dy}{dx}dx}=\int{\cos(y)dy}=\sin y$$