Mathematics - 2022-04-25

uhoh

02:44

1

Q: Upon which incorrect equation of Euler did Sophie Germain rely in her work that won a prix extraordinaire from the Paris Academy of Sciences?

Wikipedia's Sophie Germain; Work in elasticity; Subsequent attempts for the Prize says: Germain had derived the correct differential equation (a special case of the Kirchhoff–Love equation),31 but her method did not predict experimental results with great accuracy, as she had relied on an incorr...

asked in History of Science and Mathematics SE

copper.hat

03:08

why just green's functions? why not blue's or yellow's?

Ted Shifrin

Purple and chartreuse get short shrift.

copper.hat

03:36

technicolor function maybe

porridgemathematics

05:39

@TedShifrin i mean I understand 'the hermitian connection' can only be defined on a holomorphic bundle, but you can certainly define a connection on the conjugate of a holomorphic bundle - im looking at a proof of the kodaira vanishing theorem that looks at a term of the form $\nabla_{\partial_i} \nabla_{\overline{\partial_j}} (v) - \nabla_{\overline{\partial_{j}}} \nabla_{\overline{\partial_i}}(v)$ here $v$ is a $(0,1)$-form, which is a section of the conjugate of a holomorphic bundle

and the proof proceeds by writing this expression out with respect to coordinates, and curvature terms for the hermitian connection on the holomorphic tangent bundle come into play

Koro

-4

A: Real-world applications of fields, rings and groups in linear algebra.

Here's the deal, man. You know how in baby algebra, you learn a few things like how zero is the additive identity, multiplication distributes over addition, upside down fractions are multiplicative inverses? And you'll be using the real numbers, but you just think they're the normal numbers? Well...

porridgemathematics

i take your point that there is no such thing as a antiholomorphic vector bundle, it makes no sense because the composition of 'antiholomorphic' transition functions are holomorphic

Koro

Has anyone seen this answer?

leslie townes

koro: now, yes. i don't know that i've been improved by the experience

porridgemathematics

i think you can get away with defining a antiholomorphic bundle as something isomorphic to the conjugate of a holomorphic vector bundle over a complex manifold though, i dont see an immediate issue with this definition

its more or less how i was using the term

copper.hat

05:49

i am trying to unread the answer man

dearth of interesting convex questions

porridgemathematics

this is a good example of the voting system doing its job

cough.. youtube

Koro

i want to see how many downvotes it gets.

porridgemathematics

wait no, im an idiot, i need to refine what i mean by isomorphic here, because a holo. vector bundle is smoothly iso to its conjugate, sorry!

copper.hat

goodnight folks

porridgemathematics

ill try to be as precise as possible, the only situation we are talking about is the following: $X$ is a complex manifold, $(E,h)$ a hermitian holomorphic vector bundle over $X$, we restrict $E$ to a real-vector bundle over $X$, and then complexify it, we decompose this complexification via the naturally arising $J$-operator (multiplication by $\sqrt{-1}$, we need to make some choices here) into its $(1,0)$ and $(0,1)$ parts,

write $E \equiv E^{(1,0)}$ and $E \oplus \overline{E} \equiv E^{(1,0)} \oplus E^{(0,1)} = E_{\mathbb{C}}$

now there is 'the hermitian connection' $D$ on $E$, we can extend this connection to $\overline{D}$ and hence $E_{\mathbb{C}}$

and this turns out to be useful sometimes, because e.g. when $(X,g)$ is Kahler, and we take $E = T_{X}$ the holomorphic tangent bundle with the induced hermitian connection given by the Riemannian metric, then 'the hermitian connection''s extension to $T_{X,\mathbb{C}}$ coincides with the complexification of 'the Riemannian connection'

and in fact this occurs if and only if $(X,g)$ is Kahler, it characterizes the Kahler property

uh, in general we need to assume $(X,g)$ is hermitian, i.e. $g$ is invariant with respect to the $J$-operator on the real tangent bundle of $X$ to put a hermitian metric on $T_{X}$

im saying given $(X,g)$ is hermitian, its Kahler if and only if the hermitian connection (on $T_X$)'s extension to $T_X \oplus \overline{T_X}$ coincides with the complexification of the Riemannian connection to $T_X \oplus \overline{T_X}$

Prithu biswas leftmse