Let $A$ be a perfect $\kappa$-algebra over a perfect field $\kappa$ of positive characteristic $p$. Then the algebraic (= classical) cotangent complex $L_{A/\kappa}^{\operatorname{alg}}$ is known to vanish, due to the Frobenious automorphism having simultaneously to induce on the cotangent complex an automorphism and multiplication by $p$.

But we can also view $A$ and $\kappa$ as discrete $\mathbb E_\infty$-rings. The cotangent complex $L_{A/\kappa}$, which we obtain that way, is generally different from $L^{\operatorname{alg}}_{A/\kappa}$, since their homotopy groups give topological Andre-Quillen homology and (ordinary) Andre-Quillen homology respectively.

**Q:** *Can we still say something about $L_{A/\kappa}$?*

For instance:

- Does it perhaps vanish?
- Are there at least any finiteness results (e.g. when $A$ is a field, is $\dim_A \pi_n L_{A/\kappa} < \infty$)?

Perhaps a bit more broad afterquestion: *what is in general the relationship between $L_{B/A}$ and $L^{\operatorname{alg}}_{B/A}$ for a discrete commutative $A$-algebra $B$*? Other than that they coincide over the rationals and that $\pi_0$ of both is the module of Kähler differentials $\Omega_{\pi_0B/\pi_0A}$, of course.

Thanks in advance!

I am now interested in simple Lie algebras over finite fields. In Lie algebras over the complex numbers, there are several applications and some related topics.

Is there any potential application for simple Lie algebras over finite fields, or anything related? Perhaps, in coding theory or graph theory?

Let $K$ be a field, $\alpha, \beta \in \mathrm{Br}(K)$, let $X,Y$ be their Brauer-Severi Varieties, is there a way to calculate $A^*(X\times Y)$?

For example, if $\alpha,\beta$ both has degree $5$, $2\alpha=\beta$, then $A^*(X\times Y)$ is a subring (Will two non-rational equivalents cycles become rational equivalent after base change?) of $A^*(X\times Y_{\overline{K}})=A^*(\mathbb{P^4}\times\mathbb{P^4})=Z[H_1,H_2]/(H_1^5,H_2^5)$. Then $5H_i\in A^1$, $H_i\notin A^1$, but $4H_1+3H_2\in A^1$, as it is the first Chern class of the vector bundle bundle $O(1)\boxtimes (\Omega\otimes O(2))$ on $X\times Y$.

Is there a closed form for the following series?

$$2, 5, 4, 5, 6, 4, 41, 39, 3, 11, 9, 38, 7, 41, 10, 39, 8, 37, 3, 11, 40, \ldots$$

This is related to the number of odds (or 3n+1 transforms) in the Collatz sequences for the odds $3, 7, 11, 15, ...$

A general series for the number of odds (or tripling steps) in the Collatz sequence for the natural numbers is given by on OEIS, but I'm trying to make it work for these particular odds as they are very important in a paper I'm working on.

Any help or insight is much appreciated!

Suppose $A$ is an $n\times n$ real symmetric positive definite matrix. Let $A^{(-1)}_{i,j}$ be the $n\times n$ matrix the entries $(i,i),\,(i,j),\,(j,i),\,(j,j)$ of which equal to the corresponding entries of the inverse of the row $i$ column $j$ principal minor of $A$, and all other entries zero. Define $$B:=\frac{1}{n-1}\sum_{i<j}A_{i,j}^{(-1)},\quad P:=\big(\text{diag}(A)\big)^{-1},$$ and $\lambda_{\min}(M)$ to be the minimal eigenvalue of matrix $M$ (with all real eigenvalues).

Random test suggests the following inequality. Is it true? $$\lambda_{\min}(PA)\le \lambda_{\min}(BA)$$

It is true for $n=2$.

This is an equivalent and simplified reformulation of a problem in math.exchange.com with a bounty. If one finds a proof, one can reap more points and snatch the bounty there.

I tried this question twice on math.stackexchange but got no answer so I decided to move it here.

Let $M$ be a smoth manifold. Then $$C^\infty(M):=\{f:M\longrightarrow \mathbb R; f\ \textrm{is smooth}\},$$ is an $\mathbb R$-algebra with the pointwise product.

If you don't know anything about smooth manifolds it doesn't matter, all that you need to know is that $C^\infty(M)$ is an $\mathbb R$-algebra for all that follows is purely algebraic.

Let us define the space of **vector fields** on $M$ by: $$\mathfrak{X}(M):=\mathsf{der}_{\mathbb R}\ C^\infty(M):=\{X\in \mathsf{End}_{\mathbb R}(C^\infty(M)): X(fg)=fX(g)+X(f)g\}.$$ This is a $C^\infty(M)$-module with $$(f\cdot X)(g):=f X(g),$$

where $fX(g)$ is the pointwise product of the functions $f$ and $X(g)$.

We can then define the space of $p$-forms on $M$ as the $C^\infty(M)$-module:

$$\Omega^p(M):=\mathsf{Hom}_{C^\infty(M)}(\Lambda^p \mathfrak{X}(M), C^\infty(M)).$$
The **De Rham differential** on $M$ is the degree one operator $$d: \Omega^p(M)\longrightarrow \Omega^{p+1}(M)$$ given by $$\begin{eqnarray*}
d\varepsilon(X_{1}, \ldots, X_{p+1})&&:=\sum_{\sigma\in\mathsf{Sh}(1, p)} \mathsf{sgn}(\sigma) X_{\sigma(1)}(\varepsilon(X_{\sigma(2)}, \ldots, X_{\sigma(p+1)})\\
&&+\sum_{\sigma\in\mathsf{Sh}(2, p-1)}\mathsf{sgn}(\sigma) \varepsilon([X_{\sigma(1)}, X_{\sigma(2)}], X_{\sigma(3)}, \ldots, X_{\sigma(p+1)}),
\end{eqnarray*}$$ where $[\cdot, \cdot]$ is the **commutator of vector fields** which is defined by $$[X, Y]:=X\circ Y-Y\circ X.$$

**Notation:** For integers $p, q\geq 1$ let us write $S(p, q)$ as the subset of permutations $\sigma$ of the set $\{1, \ldots, p+q\}$ such that $\sigma(1)<\ldots< \sigma(p)$ and $\sigma(p+1)<\ldots< \sigma(p+q)$. The elements of $\mathsf{Sh}(p, q)$ are known as **$(p, q)$-shufles** for obvious resons.

Now, we can define a product $$\wedge: \Omega^p(M)\times \Omega^q(M)\longrightarrow \Omega^{p+q}(M)$$ setting $$(\varepsilon\wedge \eta)(X_1, \ldots, X_{p+q}):=\sum_{\sigma\in\mathsf{Sh}(p, q)} \mathsf{sgn}(\sigma) \varepsilon(X_{\sigma(1)}, \ldots, X_{\sigma(p)}) \eta(X_{\sigma(p+1)}, \ldots, X_{\sigma(p+q)}).$$ Can anyone help me to prove $$d(\varepsilon\wedge \eta)=d\varepsilon\wedge \eta+(-1)^p \varepsilon\wedge d\eta,$$ for every $\varepsilon\in \Omega^p(M)$ and $\eta\in \Omega^q(M)$?

I know the property I want to show is a classical one but I can't seem to find the proof using this algebraic formulation. I've already asked this question before and got no answer.

However, by that time things were more obscure so I decided to update with this improved version hoping someone could help me.

**Remark:**
1) The cardinality of the set $\mathsf{Sh}(p, q)$ is $\binom{p+q}{q}$. Hence we easily 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

$$\mathsf{Sh}(1, p+q)\times \mathsf{Sh}(p, q)\longrightarrow \mathsf{Sh}(p, q+1)\times \mathsf{Sh}(1, q).$$

So the problem boils do to writing those bijections explicitly.

I came to the following question when thinking about the (infinitely generated) tilting-cotilting correspondence, where it appears to be relevant.

Does there exist a locally presentable abelian category with enough injective objects that is not a Grothendieck category?

Background: an abelian category is called Grothendieck if it has a generator and satisfies the axiom Ab5. The latter means that (the category is cocomplete and) the filtered colimit functors are exact. Any locally presentable category has a generator, so the question is really about existence of locally presentable abelian categories with enough injective objects, but nonexact filtered colimits.

An abelian category is said to satisfy Ab3 if it is cocomplete. Any locally presentable category is cocomplete by definition. An abelian category is said to satisfy Ab4 if it is cocomplete and the coproduct functors are exact. Any cocomplete abelian category with enough injective objects satisfies Ab4.

The category opposite to the category of vector spaces over some fixed field is cocomplete and has enough injective objects, but it does not satisfy Ab5 (because the category of vector spaces does not satisfy Ab5*, i.e., does not have exact filtered limits). The category opposite to the category of vector spaces is not locally presentable, though.

Any locally finitely presentable abelian category is Grothendieck, and any Grothendieck abelian category is locally presentable, but the converse implications do not hold. Any Grothendieck abelian category has enough injective objects. Does a locally presentable abelian category with enough injective objects need to be Grothendieck?

My understanding of how derivations on commutative rings are like derivatives is that a derivation on $R$ is differentiation with respect to a vector field on $\text{Spec}(R)$. But derivations are supposed to be thought of as like derivatives in a wider context than commutative rings, and I don't really understand how.

Take anti-derivations on the exterior algebra of differential forms on a manifold, for instance. The exterior derivative and Lie derivatives both give you information about infinitesimal change in a differential form, but the interior derivative is defined pointwise, as an anti-derivation on the exterior algebra of the tangent space at each point, which ruins any attempt to think of anti-derivations on differential forms as capturing some information about infinitesimal change. So how can you think about interior differentiation as being like a derivative in any more concrete sense than that it obeys similar syntactic rules? More generally, how can you think about anti-derivations on exterior algebras (or more generally still, on anti-commutative graded rings) as being like a derivative?

There's also derivations on non-commutative rings. The adjoint action of an element of a ring $\text{ad}_x(y):=xy-yx$ is a derivation, but I don't see the significance of this. For example, the Pincherle derivative seems to act like a sort of "differentiation with respect to $d/dx$" insofar as it sends $d/dx$ to $1$, and the fact that it is a derivation forces certain other facts that this heutristic naively suggests to be true (for instance, that the shift operator $S_1=e^{d/dx}$ is its own Pincherle derivative). Is there some more precise way to describe the Pincherle derivative as differentiation with respect to $d/dx$? What about a way to characterize arbitrary derivations on non-commutative rings?

How about derivations on Lie algebras? The Jacobi identity can be interpreted as saying that adjoint actions are derivations, but as in the analogous fact I mentioned for derivations on non-commutative rings, I'm curious about what the significance of this is. And about how to think of arbitrary derivations on Lie algebras.

I'm trying to convert Latitude, Longitude, and Altitude values into Cartesian coordinates so I can use them as points in a Unity application I'm making. I'm using the function below. Only issue is that when I create 3d visualizations using my data, it seems to be systematically off from where they should be (if you view the image, all that data should be on relatively the same Y value. Almost as if I have to rotate the points to make it correctly correspond to its real locations. Do I need to do something more to adjust for the spherical nature of the earth? This data is only across a mile or so. I've included a picture of this. Visualization of data

/// Converts WGS84 lat/lon/height to XYZ /// References /// http://stackoverflow.com/a/8982005 /// http://mathforum.org/library/drmath/view/51832.html /// </summary> /// <returns>The WGS84 to Cartesian coordinates.</returns> /// <param name="lat">Latitude</param> /// <param name="lon">Longitude</param> /// <param name="h">height.</param> Vector3 WGS84ToCartesian(float lat, float lon, float h) { float earthRadius = 6378137.0f; float cosLat = Mathf.Cos(lat * Mathf.PI / 180.0f); float sinLat = Mathf.Sin(lat * Mathf.PI / 180.0f); float cosLon = Mathf.Cos(lon * Mathf.PI / 180.0f); float sinLon = Mathf.Sin(lon * Mathf.PI / 180.0f); float f = 1.0f / 298.257224f; float C = 1.0f / Mathf.Sqrt(cosLat * cosLat + (1 - f) * (1 - f) * sinLat * sinLat); float S = (1.0f - f) * (1.0f - f) * C; float x = (earthRadius * C + h) * cosLat * cosLon; float y = (earthRadius * C + h) * cosLat * sinLon; float z = (earthRadius * S + h) * sinLat; return new Vector3(x, y, z); }Are finite topological spaces (i.e. topological spaces whose underlying set is finite) a model for the homotopy theory of finite simplicial sets (= homotopy theory of finite CW-complexes) ?

Namely, is there a reasonable way to:

(1) given a finite topological space $X$, construct a finite simplicial set $nX$.

(2) given two finite topological spaces $X$ and $Y$, construct a simplicial set $Map(X,Y)$ whose geometric realisation is homotopy equivalent to the mapping space $Map(|nX|,|nY|)$ between the geometric realizations of $nX$ and $nY$.

(3) etc. (higher coherences)

Note: One can, of course, define $Map(X,Y)$ to be the derived mapping space between $nX$ and $nY$. But I'm wondering whether there's something more along the lines of "take the (finite) set of all continuous maps $X\to Y$ and then ... "

I know that $\sum_{i=0}^n2^i=2^{n+1}-1$. Now, let us consider $$S=2^0+2^1+2^2+2^3+\cdots$$ By a trick we have: $$S=2^0+2(1+2^1+2^2+\cdots)\Rightarrow S=1+2S$$ Therefore, $S=-1$!!

Could you please tell me what is problem? Where is the problem of this proof?

Sincerely yours.

In Davis, William J., and William B. Johnson. "Basic sequences and norming subspaces in non-quasi-reflexive Banach spaces." Israel Journal of Mathematics 14.4 (1973): 353-367., the authors discuss the following problem

"If $M$ is a norming subspace of $X^*$ and $Y$ is an $M$-closed subspace of $X$, then is $M \cap Y^\perp$ a norming subspace of $(X/Y)^*$ (where $Y^\perp$ is identified with $(X/Y)^*$ in the canonical way)?"

In the paper, the authors conclude that if $X$ is not quasi-reflexive, then there exists a norming subspace $M$ such that the conclusion fails.

What I am interested in is whether the result can be salvaged in some way by adding some reasonable conditions on $M$, and I would like to know if there is anything in the literature that has addressed this. The papers citing this paper do not discuss this particular question any further.

Thank you for your time.

Consider an elementary class $\mathcal{K}$. It is quite common in model theory that a structure $K$ in $\mathcal K$ comes with a closure operator $$\text{cl}: \mathcal{P}(K) \to \mathcal{P}(K), $$ which establishes a pregeometry on $K$.

Any pregeometry yields a notion of dimension, say:

$$\text{dim} (K) = \min \{|A|: A \subset K \text{ and } \text{cl}(A) = K\}$$ I am interested in some natural properties shared by dimensions induced by pregeometries.

What kind of properties I am looking for? An example is the following,

Suppose $K = \bigcup K_i$ (increasing chain), is it true that $\text{dim}(K) = \sup\{ \text{dim}(K_i)\}$?

I already know that there are many results like "trivial geometries are modular" but this is not the kind of result I am looking for. I am looking for structural properties of dimension because I am interested in giving an axiomatic definition of dimension.

This was asked but never answered at MSE.

Let $f(x) = \sin(a_1x) + \sin(a_2x) + \cdots + \sin(a_nx)$, where the $a_i$'s represent distinct positive integers. Suppose also that $f(x)$ satisfies the inequality $f(x) \geq 0$ on the open interval $0 < x < \pi.$

In the case n=1 of a single summand it is obvious that only $f(x) = \sin x$ satisfies the condition. For two summands, a short argument shows that only $f(x) = \sin x + \sin(3x)$ works. Following up on this, we define $g_n(x) = \sin x + \sin(3x) + \sin(5x) + \cdots + \sin((2n-1)x)$. Computing the sum explicitly yields $g_n(x) = \frac{\sin^2(nx)}{\sin x}$ which makes it clear that $g_n(x)$ is nonnegative on $(0,\pi).$

Questions: (1) Are there any other examples of $f(x)$ as above besides $g_n(x)$? If so, can one classify them all?

(2) The special case $f(x) = g_1(x) = \sin x$ satisfies the stronger condition of being strictly positive over $0 < x < \pi$. Is $\sin x$ the unique such instance of $f(x) > 0$ on $(0,\pi)$?

Thanks

In Wikipedia's page for Bertrand's postulate, it is said that its (2n,3n) version was proved by El Bachraoui in 2006. Seems likely that it was first proved way before than that! Can anyone point to the first source, or at least to a previous one?

Analogously, Wikipedia said until recently that the (3n,4n) version was due to Andy Loo in 2011. I'm aware of a proof by Denis Hanson in 1973, so I have updated the page with that info, but I don't know if his proof is the first one. Do you know of previous proofs?

Let $F(A)$ be a class of real-analytic function on an interval $A \subset \mathbb{R}$ minus the zero function.

We have the following theorem for $F(A)$.

If $f \in F(A)$ then $f$ has at most finitely many zeros $A$.

**Proof** Suppose $f\in F(A)$ has infinitely many zeros on a bounded interval. Then by Bolzano-Weierstrass the set of zeros has a convergent subsequence in $A$. Therefore, by identity theorem, $f$ must be zero on all of $A$.
However, this contradicts our assumption that $f$ is non-zero. Q.E.D.

**My question:** Are there ways of sharpening the bound on the number of zeros?

Let $N(f)$ be the number of zeros of $f$. Clear, there is no uniform bound on $N(f)$ for all $f\in F$. However, there a subset of $F$ for which we do have good upper bounds like a set of polynomials of degree $n$ in which case $N\le n$.

**My second question (or refined first question) is:** For a give $f$ which is analytic on $A$, but not a polynomial, are there ways of finding an upper bound on the number zeros?

Is every group isomorphic to an inductive colimit (that is, directed colimit, also called inductive limit, or directed limit) of free groups?

I guess, the answer is no. In that case: Is there a characterization of the groups that are isomorphic to inductive colimits of free groups?

Is every group isomorphic to an inverse limit (that is, projective limit) of perfect groups?

I guess, the answer is no. In that case: Is there a characterization of the groups that are isomorphic to inverse limits of perfect groups?