Orodruin to the rescue!
As Orodruin makes clear, a map from T to T*, (where T is the tangent space), corresponds to a tensor of type (0,2), say by rule 1), post #31, since L(T,T*) ≈ T*(tens)T*. Thus its inverse is a map from T* to T, and by the isomorphism L(T*,T) ≈ T**(tens)T ≈ T(tens)T, the...
To someone like me, i.e. me, who knows nothing of this, and never heard of Einstein's field equations before, wikipedia looks actually useful.
It is an equation between two tensors of type (0,2). (I have been calling them type (2,0)), i.e. 2-covariant tensors, linear combinations of...
Ok, so I think of it this way now: in physics there are various operations on (fields of) vectors taking them linearly or multi-linearly to other (fields of) vectors, and it is useful for calculations to represent these operations as tensor fields, hence entirely in terms of tensor products of...
Ah yes, proceeding from Orodriuin's guidance, we can see what kind of tensor the Riemann curvature should be:
We need the basic insight that the map VxV-->V(tens)V sending <v,w> to v(tens)w, is bilinear, and that thus composing with any linear map out of V(tens)V yields another bilinear map...
You are reminding me of how stunned I was when my professor asked why I was so sure, when given a function f(x), that x was the variable and f the function. Why not define x(f) to be f(x), and then f was the variable?
I.e. if F is the space of all k valued functions on the set X, then we have...
well I was enjoying the excerpt of Biennow on amazon https://www.amazon.com/Mathematical-Methods-Physics-Engineering-Mattias-ebook/dp/B086H3LMZF/?tag=pfamazon01-20
until it ran out just starting tensors, but discovered that physicists like tensors so much they use them to represent objects that...
Well a quick look at Wikipedia shows that the output of a tensor can be more general than a function, e.g. the Riemann curvature tensor ( a slight elaboration on the formula in post #21) takes a triple of vector fields and outputs a vector field.
Namely the commutator of two vector fields is...
I am a mathematician, and to me, if you already know what the tangent bundle is, and its dual the cotangent bundle, and hence know what tangent and cotangent vector fields are, then the 20 page explanation in Spivak's chapter 4, of his Comprehensive introduction to differential geometry, is as...
quantifiers are actually essential for precise statements. recall existential quantifiers affirm a statement for at least one element of a given set, while universal quantifiers do so for all such elements.
although it may sound otherwise, existential statements are often stronger than...
Excellent, except to be picky, you have omitted some existential quantifiers, e.g. in line minus 17, it should say "there exists h such that". Of course also in line -16 there is an implied universal quantifier. nice work overall.
I agree especially with points made by Choppy and CrysPhys: first of all, the purpose of writing the thesis, for most of us, is to strengthen the author's research chops to the point of being ready to do, and write up, competitive research - i.e. it is a learning experience for the author of...
When I was a young researcher (in pure mathematics), incompletely trained (by my own fault), I learned to use information I did not understand. I was happy to find out that was possible, as it helped me make research progress in less time. There was some loss of mental security since I had to...
In my opinion,
GL(n) acts, not on V, but on k^n, so it acts on the n dimensional space V only after choosing an ordered basis for V, (and thus also an ordered dual basis for V*).
Then if a is a covector in V*, represented by a row vector, and v is a vector in V, represented by a column vector...
some more detail:
let F = Q(t) where t^2 =2. then elements of F have form z = a+bt, for rational a and b. define z’ = a-bt to be the conjugate of z.
then claim if z = a +bt and w = c+dt, then z’w’ = (zw)’, i.e. the conjugate of a product of elements of F equals the product of their...