Unit tensor

In general,

(1)
\begin{align} \boldsymbol{I}= \boldsymbol{g}_i\otimes\boldsymbol{g}^i, \end{align}

i.e., for example, in 3-dim. case,

(2)
\begin{align} \boldsymbol{I}= \boldsymbol{g}_1\otimes\boldsymbol{g}^1+\boldsymbol{g}_2\otimes\boldsymbol{g}^2+\boldsymbol{g}_3\otimes\boldsymbol{g}^3 \end{align}

is called a unit tensor or a fundamental tensor.
On a surface, when $\boldsymbol{g}_1\ , \ \boldsymbol{g}_2$ are basis adjacent with a curviliniar coordinate system,

(3)
\begin{align} \boldsymbol{I}= \boldsymbol{g}_1\otimes\boldsymbol{g}^1+ \boldsymbol{g}_2\otimes\boldsymbol{g}^2 \end{align}

is also called a unit tensor. Note that, in this case, base vectors are three-dimensional vectors.
Interestingly, one can prove that

(4)
\begin{align} \boldsymbol{I}=\boldsymbol{g}^i\otimes\boldsymbol{g}_i, \end{align}

because the components of them meet. This can be verified as

(5)
\begin{align} I^{\alpha\beta}=\boldsymbol{g}^\alpha\cdot\left(\boldsymbol{g}^i\otimes\boldsymbol{g}_i\right)\cdot\boldsymbol{g}^\beta=\boldsymbol{g}^\alpha\cdot\boldsymbol{g}^\beta \end{align}

and

(6)
\begin{align} I^{\alpha\beta}=\boldsymbol{g}^\alpha\cdot\left(\boldsymbol{g}_i\otimes\boldsymbol{g}^i\right)\cdot\boldsymbol{g}^\beta=\boldsymbol{g}^\alpha\cdot\boldsymbol{g}^\beta. \end{align}

Here, we used a basic tensor calculus, i.e.,

(7)
\begin{align} \boldsymbol{a}\cdot\boldsymbol{b}\otimes\boldsymbol{c}\cdot\boldsymbol{d}=(\boldsymbol{a}\cdot\boldsymbol{b})(\boldsymbol{c}\cdot\boldsymbol{d}). \end{align}

The above two calculations are sufficient to prove that both unit tensors are substantially identical, let us check the remaining two combinations.

(8)
\begin{align} I_{\alpha\beta}=\boldsymbol{g}_\alpha\cdot\left(\boldsymbol{g}^i\otimes\boldsymbol{g}_i\right)\cdot\boldsymbol{g}_\beta=\boldsymbol{g}_\alpha\cdot\boldsymbol{g}_\beta \end{align}
(9)
\begin{align} I_{\alpha\beta}=\boldsymbol{g}_\alpha\cdot\left(\boldsymbol{g}_i\otimes\boldsymbol{g}^i\right)\cdot\boldsymbol{g}_\beta=\boldsymbol{g}_\alpha\cdot\boldsymbol{g}_\beta. \end{align}

Yes, they meet.
To summarize the above described calculations, a unit tensor has four different types of representations:

(10)
\begin{align} \boldsymbol{I}\equiv \delta_i^{\cdot j}\boldsymbol{g}^i\otimes\boldsymbol{g}_j \end{align}
(11)
\begin{align} \boldsymbol{I}\equiv \delta^i_{\cdot j}\boldsymbol{g}_i\otimes\boldsymbol{g}^j \end{align}
(12)
\begin{align} \boldsymbol{I}\equiv g_{ij}\boldsymbol{g}^i\otimes\boldsymbol{g}^j \end{align}
(13)
\begin{align} \boldsymbol{I}\equiv g^{ij}\boldsymbol{g}_i\otimes\boldsymbol{g}_j. \end{align}

Remind that a rank-2 tensor has four different types of components. Hence, Riemannian metric $g_{ij},\ g^{ij}$
and Kronecker's delta $\delta_i^{\cdot j},\ \delta^i_{\cdot j}$ can be understood as four different types of components of unit tensor.
When two tangent vectors are acting on a unit tensor from both sides, we obtain

(14)
\begin{align} \boldsymbol{a}\cdot\boldsymbol{I}\cdot\boldsymbol{b}=a_ib_jg^{ij}=a^ib^jg_{ij}=a_ib^i=a^ib_i, \end{align}

This is exactly the inner product between the tangent vectors.
Additionally, when $d\boldsymbol{x}=d\theta^i\boldsymbol{g}_i$ is acting on the unit tensor from both sides, we obtain

(15)
\begin{align} d\boldsymbol{x}\cdot\boldsymbol{I}\cdot d\boldsymbol{x}=d\theta^i d\theta^j g_{ij}. \end{align}

This is the first fundamental form.
From above consideration, it is true that

(16)
\begin{align} \boldsymbol{a}\cdot\boldsymbol{I}\cdot\boldsymbol{b}=\left<\boldsymbol{a},\boldsymbol{b}\right>. \end{align}

Hence, in a sense that it give generalized inner product and first fundamental form, it is a natural viewpoint to see that a unit tensor is essentially the Riemannian metric itself.
As the result, we see that, a unit tensor, Riemannian metric, Kronecker's delta symbol are all representing the same substance (i.e., a rank-2 tensor) but are given different representations in different contexts.

page revision: 23, last edited: 08 May 2015 04:54