@Danu I am a bit uncomfortable with this group, but U(n) sits inside Sp(2n, R) in a natural way, doesn't it? Look at the special orthogonal matrices in Sp(2n, R), eg
In arithmetic and computer programming, the extended Euclidean algorithm is an extension to the Euclidean algorithm, which computes, besides the greatest common divisor of integers a and b, the coefficients of Bézout's identity, that is integers x and y such that
a
x
+
b
y
=
gcd
(
a
,
b
)
.
{\displaystyle ax+by=\gcd(a,b).}
This is a certifying algorithm, because the gcd is the only number that can simultaneously satisfy this equation and divide the inputs. It...
@Danu In any case, $\text{Sp}(2n)$ acts transitively on the space of positive-definite symmetric matrices by $g \mapsto A^{-1}gA$. The stabilizer of the identity matrix is $\text{Sp}(2n) \cap O(2n) = U(n)$, so we have a fiber sequence $U(n) \to \text{Sp}(2n) \to \Bbb R^{n(n+1)/2}$, since the space of positive-definite symmetric matrices is convex.
Let $S$ be a formal language string over $\Sigma$, and $S_{\leqslant}$ the substring lattice. If $t$ is a substring of $S$, we write $t \leqslant S$. Define a substring ring of a string $S$ to be a unital ring $R_S$ together with a surjective map $\phi : R_S \to S_{\leqslant}$ such that:
For a...
I think I'd probably find homological algebra largely unmotivated unless I encountered the various constructions in homological algebra before in topology.
@Balarka A lot of the constructions in homological algebra turn out to be useful but are hard to have imagined first in topology. At least, I think so.
@BalarkaSen I want to verify something with you because my brain thinks they are the same thing. So suppose we have groups $G_1, ..., G_n$ and normal subgroups $H_1 ,...,H_n$. Elements of $(G_1 \times ... \times G_n) / (H_1 \times ... \times H_n)$ are of the form $(g_1,...,g_n)(h_1,...,h_n)$ right ? While elements of $(G_1/H_1) \times ... \times (G_n/H_n)$ are of the form $(g_1h_1,...,g_nh_n)$ ?
if they want people can get really mad for what i'm about to say so i will SHOUT IT LOGIC AND MATH/MATH FRAMEWORKS ARE JUST CONCEPTUAL AND SIMBOLIC LIMITS TO OUR MIND!! THE UNEXPLAINED AND THE CEASE OF OXTILITIES AGAINST UNCOMMON AND "ABNORMAL THINGS" IS THE ONLY WAY FOR EXISTENTIAL FREEDOM!! A ...
"The law that entropy always increases holds, I think, the supreme position among the laws of Nature. If someone points out to you that your pet theory of the universe is in disagreement with Maxwell's equations — then so much the worse for Maxwell's equations.
If it is found to be contradicted by observation — well, these experimentalists do bungle things sometimes. But if your theory is found to be against the second law of thermodynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation.""
Hi chat, I have a differential topology curiosity, although I admit I don't fully understand the question itself: Is the deRahm isomorphism on relative cohomology multiplicative?
If $A\subset X$, it's important to decide if you look at $k$-forms on $X$ that vanish (as forms on $X$) along $A$ or that vanish when restricted (pulled back) to $A$.
@DAnu, @Ted (this is relevant to the second space being a boundary) as far as I heard you take the differential forms that are zero on the boundary. This is apparently the same as if you took the ones that are zero on the boundary with strong regularity conditions for derivatives being zero also
@Mike I don't know if I can formulate it more concretely, it is just a curiosity that I dont really understand, however the formulation apparently was: $\rho_{rel}: \Omega^*(X,L)\to C^*(C,L)$ induces a multiplicative map on cohomology
The de Rham chain complex is special. It has the structure of a commutative differential graded algebra. This is NOT true of any of the other standard cochain complees.
@TedShifrin Take a great circle on a sphere, delete one hemisphere separated by the curve. Then the normal field of the other hemisphere along the great circle is parallel because all of them are parallel in R^3, yes? (So derivative is 0).
And you can deduce that the vector field tangent to the great circle is likewise parallel because of the constant angle. On the other hand, you can see it directly (which is what I like to emphasize geometrically) from the "physics" of constant speed circular motion.
@0celo7 They do a generalized version, relative along a map. But yes.