{title2=$\operatorname{id}_{TM}=\delta^\mu{}_{\nu}$}

The identity tensor is the canonical type-$(1,1)$ tensor that maps each tangent vector to itself. It pairs a covector $\eta$ and vector $V$ as $\eta(V)$, and its components are the <Kronecker delta> $\delta^\mu{}_{\nu}$ in every coordinate system.

Identity tensor

In <suffix notation>,
$$
\boxed{\frac{\partial x_i}{\partial x_j}=\delta_{ij}},
$$
where $\delta_{ij}$ is the <Kronecker delta>. Since $r=(x_kx_k)^{1/2}$, the <chain rule> gives
$$
\boxed{\frac{\partial r}{\partial x_j}=\frac{x_j}{r}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ia / Paper 2 / 10B / a / Solution

In coordinates $x^\mu$, write the type-$(1,1)$ tensor as
$$
T=T^\mu{}_{\nu}\,
\partial_\mu\otimes dx^\nu.
$$
For a covector $\eta=\eta_\mu dx^\mu$ and vector $V=V^\nu\partial_\nu$,
$$
T(\eta,V)=T^\mu{}_{\nu}\eta_\mu V^\nu.
$$
Since this must equal $\eta(V)=\eta_\nu V^\nu$ for every $\eta$ and $V$, the components are
$$
\boxed{T^\mu{}_{\nu}=\delta^\mu{}_{\nu}}.
$$
Thus $T$ is the <identity tensor>.

The <tensor component transformation law> gives
$$
\begin{aligned}
T^{\mu'}{}_{\nu'}
&=\frac{\partial x^{\mu'}}{\partial x^\alpha}
\frac{\partial x^\beta}{\partial x^{\nu'}}
T^\alpha{}_{\beta}\\
&=\frac{\partial x^{\mu'}}{\partial x^\alpha}
\frac{\partial x^\alpha}{\partial x^{\nu'}}\\
&=\delta^{\mu'}{}_{\nu'}.
\end{aligned}
$$
The components are therefore the same <Kronecker delta> in every coordinate system.

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 37D / a / Solution

{c}
{title2=$\delta_{ij}$}
{wiki}

The Kronecker delta is $\delta_{ij}=1$ when $i=j$ and zero otherwise. It is the component matrix of the identity map in a basis.


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