Cretu

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Let $(D,g),({\bar {D}},{\bar {g}})$ be two (Pseudo-)Riemannian charts in $\mathbb {R} ^{n}$. Let $\mathrm {f} :(D,g)\rightarrow ({\bar {D}},{\bar {g}})$ be an isommetry. Let $J_{j}^{i}(x)={\frac {\partial {}\mathrm {f} ^{i}}{\partial x^{j}}}(x)$ be the Jacobian.

We have the formula (by definition of an isommetry):

$g_{ij}(x)=(\mathrm {f} ^{*}{\bar {g}})_{ij}(x)=J_{i}^{k}(x)J_{j}^{l}(x){\bar {g}}_{kl}(f(x))$

A tensor of rank $(0,2)$ (for simplicity) (on the charts $D,{\bar {D}}$ is a collection of smooth functions ${\bar {T}}_{ij}(y),T_{ij}(x)$ that satisfy the transformation property (using a pullback $\mathrm {f} ^{*}$:

$(\mathrm {f} ^{*}{\bar {T}}_{ij})(x)={\bar {T}}_{kl}(f(x))J_{i}^{k}(x)J_{j}^{l}(x)=T_{ij}(x)$

We now define tensor densities. A tensor density of weight $W$ is a collection of smooth function $\mathbf {\bar {T}} _{ij}({\bar {g}}(y),y),\mathbf {T} _{ij}(g(x),x)$ that satisfy the transformation property:

$(\mathbf {f} ^{*}{\bar {\mathbf {T} }}_{ij})(x)={\bar {\mathbf {T} }}_{kl}({\bar {g}}(f(x)),f(x))J_{i}^{k}(x)J_{j}^{l}(x)=\det {J}^{-W}(x)\mathbf {T} _{ij}(g(x),x)=\mathbf {T} _{ij}({\bar {g}}(f(x)),x)$

Consequently, every tensor density of weight W can be written as:

$\mathbf {T} _{ij}={\sqrt {\det {g}(x)}}^{W}T_{ij}$

with an ordinary tensor $T_{ij}$.

The collection of Tensor densities ${\bar {\mathbf {T} }}_{ij}({\bar {g}}(y),y)={\bar {M}}_{ij}(y),\mathbf {T} _{ij}({\bar {g}}(f(x)),x)=M_{ij}(x)$ are thus tensors (But not over the same manifold M as g, instead it is a tensor over the section of T₂M defined by g. There is no (a priori) connection defined on that manifold. TR 09:24, 28 October 2011 (UTC)).

A tensor is a collection of multilinear maps depending only on the position x. As the dependence on $g$ is by position only, this does define a tensor as it is multilinear. Anything that looks like a tensor is a tensor. Cretu (talk) 13:17, 28 October 2011 (UTC)

Using the covariant derivative, we obtain:

$\nabla _{k}{\bar {\mathbf {T} }}_{ij}=\nabla _{k}{\bar {M}}_{ij}=\partial _{k}{\bar {M}}_{ij}+\Gamma _{ik}^{a}{\bar {M}}_{aj}+\Gamma _{jk}^{a}{\bar {M}}_{ia}={\sqrt {\det {g}}}^{W}(W\Gamma _{ak}^{a}{\bar {T}}_{ij}+\partial _{k}{\bar {T}}_{ij}+\Gamma _{ik}^{a}{\bar {T}}_{aj}+\Gamma _{jk}^{a}{\bar {T}}_{ia})$

In the next step, we will show that the definition in the article proves non well defined in connection with the Levi-Civita connection.

If we had instead:

$\nabla _{k}{\bar {\mathbf {T} }}_{ij}={\sqrt {\det {g}}}^{W}\nabla _{k}{\bar {T}}_{ij}$

we can derive a paradox. Thus this definition can not be true. Recall the property of the Levi-Civita connection:

$\nabla _{k}g(e_{i},e_{j})=g(\nabla _{k}e_{i},e_{j})+g(e_{i},\nabla _{k}e_{j})$

We consider the special case $d=1,i=j=k=1$ (i.e. one dimension):

$\nabla _{1}g(e_{1},e_{1})=2\Gamma _{11}^{1}g(e_{1},e_{1})=g_{11}g^{11}\partial _{1}g_{11}$

Question: How do you define e_i?

If these are coordinate frame fields, then they are not proper vector fields and the covariant derivative is not defined for them, and the above statement is false.

If they are proper vector fields then (det g)^1/2 != g(e₁,e₁) (The RHS is a scalar, while the LHS is not.) The are only numerically equal if e₁ coincides with the coordinate frame vector.TR 09:11, 28 October 2011 (UTC)

The e_i are the basis vectors of tangential space. $e_{i}=\partial _{i}$ (We are in euclidean charts). $g$ ist a matrix with entries $g=g_{11}=g(e_{1},e_{1})$. The determinant is $g_{11}=g(e_{1},e_{1})$. The determinant is a smooth function. Such is the right hand side. Their values are equal on the whole chart. Your mistake is that the determinant is in fact a scalar as it satisfies $\det {g}(x)=\det {g}(f^{-1}(y))$. Any arbitrary function $f$ satisfies $f=h*f'$ for suitable choices, yet it doesn't make $f$ anything else than a smooth function. Cretu (talk) 13:17, 28 October 2011 (UTC)

according to the definition of the Christoffel symbols. In 1 dimension, however:

$g_{11}={\frac {1}{g^{11}}}$

Thus:

$\nabla _{1}g(e_{1},e_{1})=\partial _{1}g(e_{1},e_{1})$

However, as we are in one dimension, we have:

$\det {g}=g_{11}=g(e_{1},e_{1})$

Thus:

$\nabla _{1}det(g)=\nabla _{1}g(e_{1},e_{1})=\partial _{1}g(e_{1},e_{1})\neq 0$

as $g_{11}$ is an arbitrary non zero smooth function. Which is an immediate contradiction to the definition in the article.