@TedShifrin Apparently one of the equivalent formulations of the second law is that for a fixed state $S_0$ of a thermodynamic system, there are arbitrarily close states $S$ in the state-space such that one cannot reach $S$ from $S_0$ by a quasi-static adiabatic path (i.e., to get from $S_0$ to $S$ you need heat exchange happening)
Basically, if $M$ is the statespace, and $\delta Q$ is the heat 1-form, then there exists points arbitrarily close to $S_0$ which are not reachable from $S_0$ by paths tangential to $\ker \delta Q$
This actually implies $\delta Q$ is integrable
(If it's non-integrable somewhere, it's totally non-integrable on a chart around that, and then you're done, because Legendrian curves approximate every curve)
But it's cool how this is just a statement about forms, right? If $\alpha$ is a 1-form on $M$ such that for any point $p \in M$ there exists points arbitrarily close to $p$ which are not reachable by paths tangent to the distribution $\ker \alpha$, then $\alpha$ is integrable
@ÍgjøgnumMeg I have some algebraic number theory questions for you
@ÍgjøgnumMeg What setting is this in? I thought there was a standard way to show that if you get isomorphisms when you localize a module homomorphism at all primes (or just at all maximal ideals) you actually started with a legitimate isomorphism.
Got it. I see why you just need maximal ideals now; I can take the maximal ideal containing the annihilator ideal of any element and then localize. The kernel of $M \to S^{-1} M$ is all the elements of $M$ which are annihilated by $S$, which would say every element of $A$ annihilates $M$.
I have something simple I think. Suppose $K$ is an algebraic number field and $\mathcal{O}_K$ be the ring of integers. I can prove every prime ideal is maximal in this following fashion: take $\mathfrak{p} \subset \mathcal{O}_K$ a prime, then $\mathcal{O}_K/\mathfrak{p}$ is a domain. Also, since $\mathcal{O}_K$ is a finitely generated $\Bbb Z$-module (under the natural incl $\Bbb Z \hookrightarrow \mathcal{O}_K$), $\mathcal{O}_K/\mathfrak{p}$ is naturally a finitely generated abelian group.
Suppose $I \subset \mathcal{O}_K$ is an ideal and I look at all the $\text{Gal}(K/\Bbb Q)$-conjugates of $I$. Let's denote these by $I^g$ for each $g \in \text{Gal}(K/\Bbb Q)$. Shouldn't it be true that $\prod I^g$ is principally generated by some integer?
I know how to do this if $K$ is imaginary quadratic. But trying to do the same thing feels like a pain
And once you have $I \subset J$ implies $I = JK$, prime ideals are indeed forced to be maximal
For imaginary quadratics, the argument is something like, if $I \subset \mathcal{O}_K$ then $I$ has a lattice basis $\{\alpha, \beta\}$ in $\Bbb C$. So $I = (\alpha, \beta)$. $\overline{I} = (\overline{\alpha}, \overline{\beta})$, and $I\overline{I} = (\alpha\overline{\alpha}, \alpha\overline{\beta}, \overline{\beta}\alpha, \beta\overline{\beta})$.
Take the gcd of the integers $\alpha\overline{\alpha}, \beta\overline{\beta}, \alpha\overline{\beta} + \overline{\beta}\alpha$. Let's say that's $n$. My claim is $I\overline{I} = (n)$.
This is some trick, say $\gamma = \alpha\overline{\beta}$. We know the coefficients of $(x - \gamma/n)(x - \overline{\gamma}/n)$ are integers, so $\gamma/n$ and $\overline{\gamma}/n$ are algebraic integers.
I attended some classes in an algebraic number theory course this semester. They proved that ring of integers of Q(zeta) is Z[zeta] for a root of unity zeta
lol Keith Conrad writes "the inclusion $\Bbb Z[\zeta_p] \hookrightarrow \mathcal{O}_{\Bbb Q(\zeta_p)}$ is an isomorphism when tensoring with $\Bbb Z_p$ for all $p$
so all ya have to do is show $\Bbb Z_p[\zeta_p] \subset \Bbb Q_p(\zeta_p)$ is the ring of integers for all $p$, which is just computations in unramified/totally ramified extensions of local fields!"
it seems all to just be "do stuff locally and stick it all together" tbh
can't wait to do more in Heidelberg :)
also $A$ is an abelian variety so
#elliptic curves
argh one of the tempo increases in Hungarian rhapsody momentarily coincided with the flashing of the cursor in this text box and for a minute I was enjoying life
Pick a non-zero rational integer $a \in I$, then $a \in \sigma_i I$ for all $i$ so it is contained in the intersection of all those guys, which is contained in the product
yeah exactly
so you want an $a$ that divides something like $\sum \prod_i \sigma_i \ell_i$
well everything simplifies when your extension is Galois because all inertial degrees and ramification indices are the same so just show that the product over conjugates of a prime is principally generated by a rational integer
and it works because the Galois group acts transitively on the primes
and norms are multiplicative
and the Galois action is multiplicative
and the norm of the prime is the guy you want as suspected
basically every ideal contains a product of powers of primes $\mathfrak{b} = \mathfrak{p}_1^{r_1} \cdots \mathfrak{p}_m^{r_m}$ so by some lemma/CRT you have $\mathcal{O}_K/\mathfrak{b} \cong \prod \mathcal{O}_K/\mathfrak{p}_i^{r_i} \cong \prod \mathcal{O}_{K, \mathfrak{p}_i}/\mathfrak{p}_i\mathcal{O}_{K, \mathfrak{p}_i}$
and then the same thing replacing $\mathcal{O}_K$ by $\mathfrak{a}$ (where $\mathfrak{a}$ is the ideal containing $\mathfrak{b}$)
but the nice thing is that the $\mathcal{O}_{K,\mathfrak{p}_i}$ are discrete valuation rings so you can immediately write $\mathfrak{a}$ as a power of the unique prime ideal
and then use the correspondence between ideals of $\mathcal{O}_K/\mathfrak{b}$ and ideals of $\mathcal{O}_K$ containing $\mathfrak{b}$ to compare the powers of primes dividing $\mathfrak{a}$ and $\mathfrak{b}$
or smth to that effect
I liked that one the most, the first proof I saw was just some ugly inclusion computations in an intro book
If you have 2 dimensional plane and you want something to be orthogonal to it lets say a line. Doesn't the line need to be in $\mathbb{R}^3$ in order to this be possible
If you think of normal vector for plane
cross product for example
orthogonal complement is set of $$ W = \{ \vec{v} \in V | \langle \vec{v}, \vec{w} \rangle = 0, \quad \forall \vec{w} \in W \} $$
Then it's expecting that you know something from the chapter that you're not using or that you already know the pattern from previous problems. But that doesn't matter to me. You just gotta do the math. You have the formula for the complement, you just need to write down what it says about the vectors.
@Misha.P i'd say yes, because its complement seems open, because, even if x is taken closer to 0, we can find tinier balls in complement of that graph with $[0,1]\times\Bbb R$ strip.
If I have subspace in $\mathbb{R}^3$ which is defined as $W^{\perp} = \begin{bmatrix} 2a \\ a \\ c \end{bmatrix}, \text{ when } a,c \in \mathbb{R}$, what kind of subspace this is?
I'm looking at a result that says that if $\|a|\\leq t$ and $a\geq 0$ then $\|a-t\|\leq t$, where $a$ is a selfadjoint element of a unital C*-algebra and $t\in\Bbb R$
Hi, can someone help me to understand why ${f(x, y, z)} ={ x^2 + y^2 − z^2}$ is a smooth map between manifolds of dimension 3 and 1. How we show that the map $f from ${R}^3$ to ${R}^1$ is a smooth map?
@AlessandroCodenotti since you solved it already i dont need to give any hints: As $a≥0$ you have $\sigma(a)\subseteq [0,\|a\|]$, then $\sigma(a-t)\subseteq [-t,\|a\|-t]$, which consists entirely of negative numbers and is bounded in absolute value by $|-t|$.
Murphy has a different proof which I quite like: look at the commutative unital C*-subalgebra generated by $1$ and $a$, using the Gelfand representation this is $C(\sigma(a))$ and then the result is obvious for real valued functions
If I have $w=\text{span}((1,-2,0))$ then orthogonal complement is on plane $w^{\perp} = \begin{bmatrix} 2a \\ a \\ c \end{bmatrix}, \quad a,c \in \mathbb{R}$ right?
@Alessandro If $D$ is a division ring, $M_n(D)$ is a simple ring, i.e., has no nontrivial 2-sided ideals. This is because if there is a nonzero ideal $I \subset M_n(D)$, it must contain a nonzero matrix $A \in I$. You can multiply by the "elementary basis matrices" $E I E'$ on the left and right to get some nonzero matrix with a single entry at the $(i, j)$-th term.
Then you can just make that an elementary basis matrix as well, with nothing anywhere except a $1$ at the $(i, j)$-th term, because you're working over a division ring
Use row and column operations to get all the elementary basis matrices with a 1 at the various diagonal positions, and finally get the identity matrix in your ideal $I$. This says $I$ is the full ring.
Wedderburn's theorem is, I think, the exact converse: say $A$ is a finite dimensional $D$-algebra for some division ring $D$ which is simple. Then $A = M_n(D)$.
i remember this specifically because a prof remarked about the theorem in a lecture, i thought hte statement was quite nice and then i used it some time later but had trouble finding a useful version of it
I get confused by varying hypothesis. One thing I know is if $A$ is simple left-Artinian then it's a matrix ring
Because grab a minimal left ideal $I \subset A$. Then $IA$ is a 2-sided ideal, forcing $IA = A$. Choose $a_1, \cdots, a_n \in A$ and $r_1, \cdots, r_n \in I$ such that $\sum r_i a_i = 1$
Look at the map $I^{\bigoplus n} \to A$ by thinking of $(r_1, \cdots, r_n)$ as a basis of $A$ over $I$. This is going to be an isomorphism, I believe.
Uh, maybe I want $n$ to be minimal, to get injectivity
Anyway now you have $A \cong I^{\oplus n}$ as left-$A$-modules, so $A = \text{End}_A(A) = \text{End}_A(I^{\oplus n}) = M_n(\text{End}_A(I))$.
So is it true that finite dimensional algebras over division rings are Artinian? Feels right to me
i dont know group cohomology, but it may also be that your formula is especially for the case $i>0$ and there may be another formulation of it that can take care of the case $i=0$ as well. But the way I read the formula I would expect $-f(g_0)$
ok if I begin reading the definition of group cohomology on wikipedia a term like $g_0- f(g_0)$ should be $g_0(0) - f(g_0)$ which is $-f(g_0)$, as $G$ should act by group automorphisms
