Math Overflow Recent Questions


The minimum of the maximum of a sequence of sinc functions

Tue, 07/03/2018 - 02:40

I apologise if this is trivial or well known to be impossible:

Can one find a finite set of integers $2\leq a_1<a_2<\ldots<a_m<\infty$ such that for the function defined as $$ f_{a_1,\ldots,a_m}(x)=\max\left\{\left|\frac{\sin (a_1 x)}{\sin x}\right| ,\left|\frac{\sin (a_2 x)}{\sin x}\right| , \cdots,\left|\frac{\sin (a_m x)}{\sin x}\right| \right\} $$ the minimum satisfies $$ \min\{f_{a_1,\ldots,a_m}(x):0\leq ~x~\leq 2\pi\}>1? $$

More generally can the lower bound be made larger? Say larger than a small positive integer? Or a slowly growing function of $m$?

How to prove the above proposition by means of number theory?

Tue, 07/03/2018 - 00:52

Function $\pi(x)$ is defined as the number of primes not greater than $x$.

Theorem 1: $\pi(x)\approx x/(2\log_{10}x)\qquad \{x|1\leqslant x\leqslant 1000, x\in \mathrm{positive~integer}\}$.

Theorem 2: $\pi(x)<k(x/\log_{10}x)\qquad \{x|1\leqslant x\leqslant+\infty,x\in \mathrm{positive~integer}\}$. Please prove that $k\geqslant 0.55$.

Theorem 3: $\pi(x)\approx k(x/\log_{10}x)\qquad \{x|1\leqslant x\leqslant +\infty, x\in \mathrm{positive~integer}\}$. Please prove that as $x$ gradually increases, although $k$ sometimes fluctuates, in general, $k$ is decreasing.

integration by parts for fractional laplacian formula for a larger class of functions

Tue, 07/03/2018 - 00:40

When is this formula valid $\int_{\mathbb R^N} (-\Delta)^sf g =\int_{\mathbb R^N} (-\Delta)^s g f $ with $s\in (0, 1)?$ I am aware of the integration by parts formula holds for $f, g$ in $L^2(\mathbb R^N)$. Is it valid for a larger class of functions which are not in $L^2$ but $\int_{\mathbb R^N} (-\Delta)^sf g$ and $\int_{\mathbb R^N} (-\Delta)^sg f$ are finite.

Also see integration by parts for the fractional Laplacian

Which open manifolds admit a transitive Lie group action?

Mon, 07/02/2018 - 23:36

Let $M$ be a connected open manifold. (i.e. non-compact and without boundary). Are there any obstructions for $M$ to admit a smooth transitive Lie group action?

(I mean a finite-dimensional Lie group).

I know that there are obstructions in the compact case.

In particular, I am interested in the case where $M$ is the space of real $d \times d$ matrices of rank bigger than $k$, for some fixed $k$. Of course, in this case there is a $\text{GL}(V) \times \text{GL}(V)$-action with finite number of orbits-each orbit corresponds to a different rank.

Even if there exists a transitive action, I don't expect it to be as natural.

$U(1)$ v.s. $SU(N)$ v.s. $SO(N)$ instantons

Mon, 07/02/2018 - 20:49

I am interested in knowing the details of the comparison between $U(1)$, $SU(N)$ and $SO(N)$ instantons for their gauge theories in 4 spacetime dimensions., in terms of:

  1. Chern class (1st, 2nd), and

  2. Pontryagin class

  3. And their normalization and mapping between the instanton numbers. For example, the unit instanton number 1 in $SO(3)$ gauge theory may map to the instanton number 1/4 in $SU(2)$ gauge theory on a non-spin manifold, while the unit instanton number 1 in $SO(3)$ gauge theory may map to the instanton number 1/2 in $SU(2)$ gauge theory on a non-spin manifold etc.

More precisely, can we compare the quantization of the charcteristic classes $$ p_1(SO(N)), $$ $$ c_2(SU(N)=\frac{1}{8 \pi^2}\int \text{tr}(F_{(SU(N))}\wedge F_{(SU(N))}), $$ $$ c_1^2(U(1))=\int \frac{F_{U(1)}}{2 \pi} \wedge \frac{F_{U(1)}}{2 \pi} =\frac{1}{4 \pi^2} \int {F_{U(1)}} \wedge {F_{U(1)}} , $$ on the spin manifolds and non-spin manifolds?

We can also embed the $U(1) \subset SU(N)$, $U(1) \subset SO(N)$ and $SO(N) \subset SU(N)$ to compare the instanton numbers of $G_a$ evaluated as the instanton numbers of $G_b$ whenever $G_a \subset G_b$. So how are the instanton number maps from that of $G_a $ to that of $ G_b$?

Here $F=dA + A \wedge A$ is the 2-form curvature of the 1-form connection $A$.

Genus of Cayley graph of $A_5$ with two generators of order 5

Mon, 07/02/2018 - 19:36

The Cayley graph of $A_5$ with two generators of order 5 (I believe all such are isomorphic) seems rather complicated. I am wondering what is its graph genus (orientable or non-orientable). The best I could get by trial and error is an embedding without crossings on a sphere with 10 crosscaps:

Profiles of very high dimensional functions

Mon, 07/02/2018 - 19:08

This question comes from trying to understand the recent success of deep neural nets. Neural networks just (crudely speaking) create a very complicated function of very many variables, and then perform gradient descent. It would be natural to expect them to get stuck in local minima, but that does not seem to be a serious problem. The practitioners, when asked, seem to believe that this is because in very high dimensions it is very unlikely that a critical point is a local minimum (since the signs of the eigenvalues of the Hessian are, in some sense, randomly distributed, so, since the dimension is high, it is unlikely that they are all negative). Is there any notion of "random functions" that makes this reasoning, well, reasonable? Presumably, similar phenomena should occur in mathematical physics...

Create "new" Vassiliev knot invariants?

Mon, 07/02/2018 - 17:25

If I want to create ("my own") finite type knot invariant, how would I approach it? Can I just pick a degree and in some process figure out the definition of the invariant?

Prove that for any two rational numbers r and s with r ≠ 0 , 1/ (r − s) is rational [on hold]

Mon, 07/02/2018 - 17:21

Hello I have the following to problem

Prove that for any two rational numbers r and s with r ≠ 0 , 1/ (r − s) is rational.

My proof process is as follows (may be incorrect)

Suppose r and s are rational numbers.

By definition of rational, they can be expressed as a quotient of two integers with a nonzero denominator.

Then 1/r is equivalent to (1- sr)/r

Not sure where to go after this step

Coefficients $U_m(n,k)$ in the identity $n^{2m+1}=\sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot n^k$

Mon, 07/02/2018 - 17:07

Review the main result of, that is the identity \begin{equation}\label{f1} n^{2m+1}=\sum\limits_{1\leq k \leq n}\sum\limits_{j\geq0}A_{m,j}k^j(n-k)^j, \end{equation} where $A_{m,j}$ is from sequences A302971 and A304042. In this question we discuss the polynomial \begin{equation}\label{f2} \sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot n^k, \ m\geq0 \ \mathrm{integer} \end{equation} That is generated by the identity \begin{equation}\label{f3} (1.3)\quad\sum\limits_{1\leq k \leq T}\sum\limits_{j\geq0}A_{m,j}k^j(n-k)^j=\sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot n^k, \end{equation} where $T=1,2,3...$ and $m\geq0, \ m=\mathrm{const}$. The coefficient $A_{m,j}$ is generated by \begin{equation*}\label{gen_13} A_{m,j}:= \begin{cases} 0, & \mathrm{if } \ j<0 \ \mathrm{or } \ j>m \\ (2j+1)\binom{2j}{j} \sum_{d=2j+1}^{m} A_{m,d} \binom{d}{2j+1} \frac{(-1)^{d-1}}{d-j} B_{2d-2j}, & \mathrm{if } \ 0 \leq j < m \\ (2j+1)\binom{2j}{j}, & \mathrm{if } \ j=m \\ \end{cases} \end{equation*} Derivation of coefficients $A_{m,j}$ is discussed in In particular, the right part of (1.3) returns odd power $2m+1$ of $T\in\mathbb{N}$ when $n=T$ \begin{equation*} T^{2m+1}=\sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot T^k \end{equation*} We denote the part $\sum\nolimits_{j\geq0}A_{m,j}k^j(n-k)^j$ of the left part of equation (1.3) as \begin{equation}\label{f4} D_m(n,k)=\sum\limits_{j\geq0}A_{m,j}k^j(n-k)^j \end{equation} Therefore, below we attach a tables containing the polynomials $\sum\nolimits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot n^k$ for various $t$ and $m$, that is shown in each table's caption. The main question is:

Question 1. Is there a recurrent that gives the coefficients $U_m(n,k)$ otherwise then by the identity \begin{equation}\label{f3_1} \sum\limits_{1\leq k \leq T}\sum\limits_{j\geq0}A_{m,j}k^j(n-k)^j=\sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot n^k, \end{equation} i.e is there any function $F(n,m)$ such that $F(m,n)=U_m(n,k)$? The following arrays could be generated using Mathematica code Um(n,k)_coefficients2.txt. Here we begin to show our arrays of polynomials for $m=1,2,3,4$ and $T=1,2,...,10$

Array for $m=1$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) T=1 &:& -5 + 6 n \\ T=2 &:& -28 + 18 n \\ T=3 &:& -81 + 36 n \\ T=4 &:& -176 + 60 n \\ T=5 &:& -325 + 90 n \\ T=6 &:& -540 + 126 n \\ T=7 &:& -833 + 168 n \\ T=8 &:& -1216 + 216 n \\ T=9 &:& -1701 + 270 n \\ T=10 &:& -2300 + 330 n \end{eqnarray*} Coefficients of above polynomials are terms of sequences A028896 and A275709.

Array for $m=2$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) T=1 &:& 31 - 60 n + 30 n^2 \\ T=2 &:& 512 - 540 n + 150 n^2 \\ T=3 &:& 2943 - 2160 n + 420 n^2 \\ T=4 &:& 10624 - 6000 n + 900 n^2 \\ T=5 &:& 29375 - 13500 n + 1650 n^2 \\ T=6 &:& 68256 - 26460 n + 2730 n^2 \\ T=7 &:& 140287 - 47040 n + 4200 n^2 \\ T=8 &:& 263168 - 77760 n + 6120 n^2 \\ T=9 &:& 459999 - 121500 n + 8550 n^2 \\ T=10 &:& 760000 - 181500 n + 11550 n^2 \end{eqnarray*}

Array for $m=3$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) T=1 &:& -125 + 406 n - 420 n^2 + 140 n^3\\ T=2 &:& -9028 + 13818 n - 7140 n^2 + 1260 n^3\\ T=3 &:& -110961 + 115836 n - 41160 n^2 + 5040 n^3\\ T=4 &:& -684176 + 545860 n - 148680 n^2 + 14000 n^3\\ T=5 &:& -2871325 + 1858290 n - 411180 n^2 + 31500 n^3\\ T=6 &:& -9402660 + 5124126 n - 955500 n^2 + 61740 n^3\\ T=7 &:& -25872833 + 12182968 n - 1963920 n^2 + 109760 n^3\\ T=8 &:& -62572096 + 25945416 n - 3684240 n^2 + 181440 n^3\\ T=9 &:& -136972701 + 50745870 n - 6439860 n^2 + 283500 n^3\\ T=10 &:& -276971300 + 92745730 n - 10639860 n^2 + 423500 n^3 \end{eqnarray*}

Array for $m=4$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) T=1 &:& 751 - 2640 n + 3780 n^2 - 2520 n^3 + 630 n^4\\ T=2 &:& 162512 - 325440 n + 245700 n^2 - 83160 n^3 + 10710 n^4\\ T=3 &:& 4297023 - 5837040 n + 3001320 n^2 - 695520 n^3 + 61740 n^4\\ T=4 &:& 45586624 - 47125200 n + 18484200 n^2 - 3276000 n^3 + 223020 n^4\\ T=5 &:& 291683375 - 244000800 n + 77546700 n^2 - 11151000 n^3 + 616770 n^4\\ T=6 &:& 1349845776 - 949440240 n + 253906380 n^2 - 30746520 n^3 + 1433250 n^4\\ T=7 &:& 4981676287 - 3024769440 n + 698619600 n^2 - 73100160 n^3 + 2945880 n^4\\ T=8 &:& 15551330048 - 8309593440 n + 1689523920 n^2 - 155675520 n^3 + 5526360 n^4\\ T=9 &:& 42670773999 - 20362676400 n + 3698370900 n^2 - 304479000 n^3 + 9659790 n^4\\ T=10 &:& 105670786000 - 45562677600 n + 7478370900 n^2 - 556479000 n^3 + 15959790 n^4 \end{eqnarray*} The PDF-analog of this question with extended data of $U_m(n,k)$ coefficients up to $T=40$ is available at this link.

relax a rectangular linear assignment problem

Mon, 07/02/2018 - 16:47

I wonder if there is any literature on the following problem

$\min_{X\in R^{m\times n}}\sum_{i,j} C_{i,j}X_{i,j}, s.t. \sum_{i}X_{i,j}=\sum_{j}X_{i,j}=1, X_{i,j}\geq 0$

The closest related problem might be Rectangular Linear Assignment Problem (RLAP), as RLAP further constrains $X_{i,j}$ to take only 0 or 1 (refer to I understand that the proposed problem is a relax version of the RLAP. But my intuition is that the optimum for the relax problem should occur at "vertex". So, do the relaxed RLAP share the same optimum as RLAP?

Tarski number is not a quasi isometric invariant, an example?

Mon, 07/02/2018 - 16:28

I know that Tarski number is not a quasi isometric invariant, i.e.

Let $G,H$ be two groups such that $G\sim_{QI} H$, then it is not necessary to have $T(G)=T(H)$.

But can you bring an example for this statement?!

Pushout of spaces

Mon, 07/02/2018 - 15:06

Suppose that we have a map between two pushout diagrams of topological spaces $$ [A_{1}\leftarrow A_{0}\rightarrow A_{2} ] \rightarrow [B_{1}\leftarrow B_{0}\rightarrow B_{2} ]$$ such that for any $i\in\{0,1,2\}$, $A_{i}\rightarrow B_{i}$ is a trivial cofibration of topological spaces. And suppose that all spaces are CW-complexes. Moreover all maps $A_{i}\rightarrow A_{j}$ and $B_{i}\rightarrow B_{j}$ are supposed to be surjective. What can we say about the induced map $$ colim [A_{1}\leftarrow A_{0}\rightarrow A_{2} ] \rightarrow colim[B_{1}\leftarrow B_{0}\rightarrow B_{2} ]$$ e.g. is it a weak equivalence ?

dataset imbalancing error in R [on hold]

Mon, 07/02/2018 - 13:56

The dataset that I'm working on is unbalanced, so I'm trying to balance the dataset by using undersampling but I get an error

library(ROSE) data_frame <- click.csv data_frame2 <- buy.csv colnames(data_frame) [1] "Session ID" "Timestamp" "Item ID" "Category" colnames(data_frame2) [1] "Session ID" "Timestamp" "Item ID" "Price" "Quantity" mydata<- merge(x=data_frame, y=data_frame2, by = "SessionID", all.x = TRUE, allow.cartesian=TRUE)# left outer join mydata >mydata Session ID Timestamp.x Item ID.x Category Timestamp.y Item ID.y Price Quantity 1: 1 2014-04-07T10:51:09.277Z 214536502 0 <NA> <NA> <NA> <NA> 2: 1 2014-04-07T10:54:09.868Z 214536500 0 <NA> <NA> <NA> <NA> 3: 1 2014-04-07T10:54:46.998Z 214536506 0 <NA> <NA> <NA> <NA> 4: 1 2014-04-07T10:57:00.306Z 214577561 0 <NA> <NA> <NA> <NA> 5: 10000001 2014-09-08T10:35:38.841Z 214854230 S <NA> <NA> <NA> <NA> --- 40596049: 9999997 2014-09-07T18:12:46.466Z 214854159 S <NA> <NA> <NA> <NA> 40596050: 9999997 2014-09-07T18:13:04.315Z 214643036 S <NA> <NA> <NA> <NA> 40596051: 9999997 2014-09-07T18:14:47.365Z 214854159 S <NA> <NA> <NA> <NA> 40596052: 9999998 2014-09-07T20:53:43.120Z 214541597 0 <NA> <NA> <NA> <NA> 40596053: 9999999 2014-09-04T04:44:46.942Z 214644650 S <NA> <NA> <NA> <NA> mydata$ItemID.y[!$ItemID.y)]<-1 mydata$ItemID.y[$ItemID.y)]<-0 table(mydata$ItemID.y) 0 1 29698257 10897796 > str(mydata) Classes ‘data.table’ and 'data.frame': 40596053 obs. of 8 variables: $ SessionID : Factor w/ 9249729 levels "1","10000001",..: 1 1 1 1 2 2 2 2 2 3 ... $ Timestamp.x: Factor w/ 32937845 levels "2014-04-01T03:00:00.124Z",..: 1406509 1407501 1407712 1408409 29083768 29085345 29085440 29085649 29088238 29247009 ... $ ItemID.x : Factor w/ 52739 levels "1178793047","1178794001",..: 2083 2082 2084 9906 50230 6411 6410 50230 50187 48852 ... $ Category : Factor w/ 339 levels "0","1","10","11",..: 1 1 1 1 339 339 339 339 339 339 ... $ Timestamp.y: Factor w/ 1136477 levels "2014-04-01T03:05:31.743Z",..: NA NA NA NA NA NA NA NA NA NA ... $ ItemID.y : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ... $ Price : Factor w/ 735 levels "0","10052","1015",..: NA NA NA NA NA NA NA NA NA NA ... $ Quantity : Factor w/ 28 levels "0","1","10","11",..: NA NA NA NA NA NA NA NA NA NA ... - attr(*, ".internal.selfref")=<externalptr> data_balanced_over <- ovun.sample(ItemID.y ~ ., data = mydata, method = "over",N = 800) here is the error that I got Error in function (formula, data, method, subset, na.action, N, P=0.5, :

The response variable has only one class.

how can solve this error?

Bi-Lipschitz extension

Mon, 07/02/2018 - 13:43

Given a bi-Lipschitz homeomorphism $\Phi:\mathbb{B}^n(0,1)\to\mathbb{R}^n$, (that is a bi-Lipschitz map onto the image), can one find a bi-Lipschiz homeomorpism $\Psi:\mathbb{R}^n\to\mathbb{R}^n$ such that $$ \Psi|_{\mathbb{B}^n(0,\frac{1}{2})}=\Phi|_{\mathbb{B}^n(0,\frac{1}{2})}\ \ ? $$

Algorithms for Balanced Coloring of Complete Symmetric Graphs

Mon, 07/02/2018 - 05:26


Has this the following problem already been studied:

given a complete, weighted, finite symmetric Graph $G$ with $kn$ vertices, assign to each of the vertices one color from $k$ different colors, so that each vertex along with its $k-1$ nearest neighbors represents all $k$ colors.

So far I could only find something about Equitable Coloring or the relaxed version, which is called Balanced Coloring. Both of these are related to topological graphs and the objective is different.

I already managed to come up with some heuristics, but would like to know what the state of knowledge about that problem is and where it naturally arises.

A note on resolving ambiguity (in reply to Brendan McKay's comment):
the comparison of edges is assumed to be based on their indices in a fixed sequence, that is sorted according to non-decreasing edge-lengths, i.e. if we have two edge $e_{ij}$ and $e_{jk}$ then $v_i$ is nearer to $v_j$ than is $v_k$ iff either both edgeweights are equal and $e_{ij}$ comes before $e_{jk}$ in the sorted sequence or else, if the weight of $e_{ij}$ is smaller than that of $e_{jk}$ .


A question on a proof in the Ralph Greenberg's paper "On a Certain l-Adic Representation"

Sun, 07/01/2018 - 18:54

I'm currently reading the paper "On a Certain l-Adic Reprersentation" written by Ralph Greenberg.(Inventiones 1973) And I'm stuck with a proof of the Proposition 2.

Here $k$ is a totally imaginary abelian extension of $\mathbb{Q}$, $K/k$ is the cyclotomic $\mathbb{Z}_{l}$-extension. $A_{n}$ is the $l$-primary subgroup of the ideal class group of $k_{n}$.

First of all, I don't know why there exists a "root of unity" $w$ in $k_{n}$ such that $N_{n,0}(\beta)=N_{n,0}(w)$.

Instead, Using the condition $c \in A_{n}^{-}$, and as the complex conjugation $J$ on $k_{n}$ acts on $\sigma$ (a generator of the galois group $G(k_{n}/k)$) by $J \sigma J = {\sigma}^{-1}$. We can see that there exists a "unit" $u$ and an element $\gamma$ such that $\beta = u \gamma^{\sigma -1}$.

Secondly, I don't understand the line "If $b=a^{1-J}$, then $b^{\sigma - 1}=(\gamma)^{\sigma -1}$."

Here is my computation (that might be wrong.)

$b=a \overline{a}^{-1}$

As $c \in A_{n}^{-}$, there exists $\delta \in k_{n}$ such that $a \overline{a}=(\delta)$.

$\sigma(a) \sigma(\overline{a}) = (\sigma(\delta))$.

$b^{\sigma -1} = \sigma(a) \overline{a} \sigma(\overline{a})^{-1} a^{-1} = (\sigma(a) a^{-1})^{2} (\delta^{1-\sigma})=(\alpha)^{2}(\delta^{\sigma-1})$.

For the proof of proposition 2 to be true, we need $\alpha^{2}$ must be a unit times an element of the form $\phi^{\sigma-1}$. But I don't see why.

An inequality related to Riesz–Thorin theorem, determinants and $L_p$ norm

Sun, 07/01/2018 - 18:33

Let $a, b, c \in \mathbb{R}^n$ , $p \in [1, +\infty)$, prove that

$$\left( \sum_{1\leq i < j <k \leq n} \left| \det\left(\begin{matrix} a_i & b_i & c_i \\ a_j & b_j & c_j \\ a_k & b_k & c_k \end{matrix}\right)\right|^p \right)^{\frac{1}{p}} \leq c_p \left( \sum_{i=1}^n |a_i|^p \right)^{\frac{1}{p}} \left( \sum_{1\leq j <k \leq n} \left| \det\left(\begin{matrix} b_j & c_j \\ b_k & c_k \end{matrix}\right)\right|^p \right)^{\frac{1}{p}} $$

where $c_p = \max(1, 3^{1-\frac{2}{p}})$.

A 2-dimensional analogue of this problem was discussed here: An inequality related to Lagrange's identity and $L_p$ norm


  1. When $p=1$, the proof is straightforward since $$\left| \det\left(\begin{matrix} a_i & b_i & c_i \\ a_j & b_j & c_j \\ a_k & b_k & c_k \end{matrix}\right)\right| \leq |a_i| \left| \det\left(\begin{matrix} b_j & c_j \\ b_k & c_k \end{matrix}\right)\right| + |a_j| \left| \det\left(\begin{matrix} b_i & c_i \\ b_k & c_k \end{matrix}\right)\right| +|a_k| \left| \det\left(\begin{matrix} b_i & c_i\\ b_j & c_j \end{matrix}\right)\right| $$, by Laplace expansion and triangle inequality. Summing up all these inequalities is enough. $p = \infty$ case can be proved in a similar way.

Using Holder's inequality directly on the Laplace expansion gives a weaker bound: $3^{1 - \frac{1}{p}}$

  1. When $p=2$, LHS is the volume of Parallelepiped spanned by three vectors $a,b,c$, while RHS is the norm of $a$ times the area of parallelogram spanned by $b,c$, so the inequality is clearly true. (This fact can be proved by using Cauchy-Binet)

  2. As users @fedja and @mahdi suggested in An inequality related to Lagrange's identity and $L_p$ norm , this problem is closely related to Riesz-Thorin interpolation theorem. However, I find it difficult to apply the theorem directly on my problem.


Is the category of enriched operads (co)complete?

Sun, 07/01/2018 - 10:34

Let $V$ be a symmetric monoidal category which is complete and cocomplete. Is the category of small symmetric colored $V$-enriched operads complete and cocomplete? If $V$ is presentable, is it presentable?


In the case of $V$-enriched categories, corresponding results are proven in the following papers:

  • Wolff, $V$-cat and $V$-graph.
  • Kelly, Lack, $V$-Cat is locally presentable or locally bounded if $V$ is so.

Functoriality for wrong way maps

Sun, 07/01/2018 - 10:01

In the K-theory formulation of the index theorem one defines the topological index in terms of the so called wrong way maps. Those maps are defined for embeddings of compact manifolds $i:X \to Y$: see section 2.2 (page 16) in Landweber - K-theory and elliptic operators article for more details. I would like to understand

why this construction is functorial, i.e. why $(j \circ i)_!=j_! \circ i_!$ where $i:X \to Y$ and $j:Y \to Z$ are embeddings of compact manifolds.

Forgive me if this question is too elementary for this site.