On Mon, Nov 24, 2025 at 03:33:04AM -0800, Alan Grayson wrote: > > > On Sunday, November 23, 2025 at 4:17:45 PM UTC-7 Russell Standish wrote: > > On Sat, Nov 22, 2025 at 03:59:00AM -0800, Alan Grayson wrote: > > > > > > On Friday, November 21, 2025 at 3:27:17 PM UTC-7 Russell Standish wrote: > > > > On Thu, Nov 20, 2025 at 07:02:53PM -0800, Alan Grayson wrote: > > > > > > > > > I studied linear algebra, but my questions involve tensors. If a > tensor > > > T is defined as a linear function whose domain is a vector space, and > > > maps to a real number, how does one get a real number from T(u), if we > > > do the calculation using matrices? Here there is no v, just u. AG > > > > A matrix corresponds to a rank 2 tensor, ie T(u,v)∈R. T(u)∈R > > corresponds to a rank 1 tensor. In matrix notation, a rank 1 tensor is > > a transposed vector, ie vᵀ for some vector v∈Rⁿ. vᵀu in matrix > > notation corresponds to v.u (ie dot or inner product of two vectors). > > > > > > I'm seeking an unambiguous definition of a TENSOR. You wrote earlier > > that a tensor is a MAP whose arguments are VECTORS in a vector space, > > which MAP to real numbers, and is INVARIANT under changes in coordinate > > systems. Your definition seems OK, but upon more analysis I find it > > kind-of vacuous. Firstly, any function which depends on elements in a > > vector space which are invariant under changes in coordinate systems, > > will necessarily be invariant under changes in coordinate systems, and > > it doesn't matter if that function is linear or not in its arguments. > So, is > > a tensor just limited by the condition of linearity of its arguments? > The > > invariance under coordinate transformations is a direct result of what > > its arguments are, and since vectors are invariant, so the tensor T must > > also have this property. That is, the invariance property of T is > totally > > dependent on the invariance property of its domain, the invariant > vectors > > in some vector space. TY, AG > > Correct so far. Yes - it seems trivial so far, but when you work out > how the components of the tensor change with changes of the > coordinate system you get the concept of covariance, and when you > apply the tensors to tangent spaces on Riemann manifolds, it become > decidedly non-trivial. When you're ready, you'll need to work through > coveriant differentiation, where there is an additional term coming > from curvature of spacetime. > > > So far, the concept of tensors seems unrelated to the additional term > you allege that relates to differentiating the tensor field. Can you say > something more informative about this result? AG >
In a handwavy way, the additional term (aka Christoffel symbol) comes from the fact that the coordinate system itself must change as the tangent space changes from point to point in a Riemann manifold. Of course, for the details, you need to work out how to translate tangent spaces, which involve full on tensor calculus. From memory, Misner, Thorne and Wheeler give a pretty good account of how to do that. > > > Moreover, you claim an invariant vector is in fact a tensor of rank 1. > > It is actually the transpose of a vector. > > > What is its MAP? > > > > The transpose uᵀ. Using tensor multiplication, uᵀv === u.v, where . is > the familiar inner product. > > > Why do you need to introduce v to evaluate T(u), > > v is the vector corresponding to the tensor T. > > > No. v has nothing to do with T(u), unless you explicitly define v as uᵀ. > I suspect what your definition of tensors is lacking is any reference or > use of the dual space related to the vector space on which the domain > of T operates. AG > Sorry - I switched notation in the previous paragraph. I think my original post had T(u) = vᵀu - so v is the vector corresponding to T: more accurately vᵀ is the tensor, and it is a vector in the dual space. The terms covariant vectors (for the dual space) and contravariant vectors (for the original tangent space) are also used, to highlight how the components of these things change with coordinate system change. Since dual spaces are also vector spaces, one can equally talk about the original vector u being a tensor, this time representing a map from the dual space to the field (ie the reals). -- ---------------------------------------------------------------------------- Dr Russell Standish Phone 0425 253119 (mobile) Principal, High Performance Coders [email protected] http://www.hpcoders.com.au ---------------------------------------------------------------------------- -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/aSUC9u0_ZlG6aRuR%40zen.
