The scalar no scalar mystery

The scalar no scalar mystery. 11.22.15 Here is the discussion on this subject , _______________________________________________________________ Scalar under transformations Imagine a version of Picture A (2.1.1) where we transform from x-space with basis vectors ei to some x'-space with basis vectors e'i. Under such a transformation, a dot product like α v transforms as a scalar. So, whereas v and α transform as vectors, the function α(v) transforms as a scalar field, just as in (2.2.6). That is to say, α(v) = α v = α' v' = α'(v') . (2.11.13) Although the function α(v) transforms as a scalar, the associated vector α (covector) transforms as a rank-1 tensor (vector), and it is through this isomorphic connection that we loosely refer to the function α(v) as being a rank-1 tensor. When functional α is expanded on the basis functionals λi, the coefficients αi are the rank-1 tensor components, and these coefficients are projected out by (2.11.9). ____________________________________________________________________ In App D tensor functions are somehow not scalars. What does App D say about a vector functional? <T | vi> = T(vi) Ti = T(ei) MIJ = (vj)i T(vi) = Σj (vi)j Tj α(vi) = Σj (vi)j αj α(vi) = Σj (vi)j αj αj = corresponding rank-1 tensor (vector) = vi α // there it is, finally; sure looks like a "scalar". α(v) = Σj vj αj αj = corresponding rank-1 tensor (vector) = v α // there it is, finally; sure looks like a "scalar". Jump ahead to my transformation rule applied to a rank-1 tensor function. α'(v1) = α( R1jvj) = α(v'1) = R1jα(vj) v1 = R1beb Well the equation can be selected to be α(v'i) = Rijα(vj) = α'(vi) v'i = Rij vj α' Where is "scalar" now? App D says this is the transformation rule for a rank-1 tensor function. HERE, Rij are the lincom transformation coefficients which generate v'i from vi . THAT is the underlying transformation being talked about in my App D interpretation. Side by side: α(vi) = α vi = α' vi' = α'(vi') v'i = Rij vj Chapter 2 α'(vi) = α(v'i) = Rijα(vj) v'i = Rij vj Appendix D Comment: The first line uses some component transformation (v'i)a = Rab (vi)b (α'i)a = Rab (αi)b and the second line use a vector transformation, so that explains the mystery. What happens at the k = 2 level? T(v'i,v'i) = RijRijT(vj,vj) T(v1,v2) = Σab Tab (v1)a(v2)b Under a coordinate transformation you would say T'(v'1,v'2) = Σab Tab (v1')a(v2')b (v1')a = Rar(v1)r (v2')b = Rbs(v2)s T'ab = RadRbeTde // these three lines are all component transformations Then T'(v'1,v'2) = Σab T'ab (v1')a(v2')b = RadRbeTde Rar(v1)r Rbs // all implied sums = RadRbeRarRbs (v1)r(v2)s = (RadRar)(RbeRbs) Tde(v1)r(v2)s = δdrδesTde(v1)r(v2)s =Tde(v1)d(v2)e =Tab(v1)a(v2)b = T(v1,v2) = = RabRbc Σab Tab(v1)b (v2)c D.4 Two kinds of transformations