I understand the $2^{nd}$ order tensor is the same as the common Identity matrix of $2^{nd}$ order and can be written as $${\bf I}=\delta_{ij} {\bf{e}}_i \otimes {\bf{e}}_j$$ Which can be translated to $${\bf I}=\delta_{11} {\bf{e}}_1 \otimes {\bf{e}}_1+\delta_{12} {\bf{e}}_1 \otimes {\bf{e}}_2+\delta_{21} {\bf{e}}_2 \otimes {\bf{e}}_1+\delta_{22} {\bf{e}}_2 \otimes {\bf{e}}_2$$
where,$e_1= \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $e_2= \begin{bmatrix} 0 \\ 1 \end{bmatrix}$ The dyad operator is defined as $\bf a \otimes b = a b^T$
The $4^{th}$ order Identity tensor is given as $${\mathbb{I}}=\delta_{ij} \delta_{kl} {\bf{e}}_i \otimes {\bf{e}}_j \otimes {\bf{e}}_k \otimes {\bf{e}}_l$$ However, I do not understand the $4^{th}$ order Identity tensor which involves dyadic product between matrices. I mean in a computationally amenable form. I understand it is very basic, but to the best of my knowledge I could not find the physical form (matrix representation) of this $4^{th}$ order tensor, $\mathbb{I}$. I am learning elasticity from an implementation point and so I thought understanding basics of tensors would help. If this is already available in the internet please do not close the question. Just share the link to the answer. Also refer some books/documents that explain tensor operations computationally.
$ \def\a{\alpha}\def\d{\delta}\def\e{\varepsilon} \def\s{\star}\def\o{\otimes} \def\p{\partial} \def\A{{\cal A}}\def\B{{\cal B}}\def\C{{\cal C}} \def\E{{\cal E}}\def\F{{\cal F}}\def\G{{\cal G}} \def\I{{\mathbb I}}\def\S{{\cal S}} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\trace#1{\op{tr}\LR{#1}} \def\sym#1{\op{sym}\LR{#1}} \def\qiq{\quad\implies\quad} \def\qif{\quad\iff\quad} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} \def\c#1{\color{red}{#1}} \def\CLR#1{\c{\LR{#1}}} \def\fracLR#1#2{\LR{\frac{#1}{#2}}} $While common in Elasticity and Continuum Mechanics, the symbol $\o$ is used for the Kronecker product rather than the dyadic product in most other fields.
This is because those other fields rarely require a dyadic product, but the Kronecker product is extremely useful for flattening matrices into vectors $\,($and tensors into matrices, thereby avoiding tensors altogether$)$.
I would also mention that in physics, simple juxtaposition is often used for dyadics, i.e. $$a\o b\o c \implies abc$$ A better choice is Index Notation, which is utilized in all fields. It's also clear and compact, especially if you omit the superfluous cartesian basis $\c{\rm vectors}$ $$a\o b\o c = a_ib_jc_k\CLR{e_i\o e_j\o e_k} \qiq a_ib_jc_k\\$$
Now to answer your question, there are three different isotropic fourth-order tensors, all of which can be expressed in terms of the Kronecker delta symbol $$\eqalign{ \d_{jk}\d_{\ell m} = \G_{jk\ell m} = \F_{j\ell mk} = \E_{j\ell km} }$$ Note that $\G$ is simply the dyadic product of two identity matrices, while $\E$ and $\F$ are obtained by permuting the indices. Every component of these tensors is $\,\in\{0,{\tt1}\}$
Just as the single dot product is the sum over a single adjacent index, the double-dot product is the sum over an ordered pair of adjacent indices, e.g. $$\eqalign{ \def\sk{\sum_{k=1}^m} \def\sl{\sum_{\ell=1}^n} C &= A\cdot B \qiq C_{ij} = \sk A_{i\c{k}}\,B_{\c{k}j} \\ \C &= \A:\B \qiq \C_{ijpq} = \sk\sl \A_{ij\c{k\ell}}\,\B_{\c{k\ell}pq} \\ M &= N:\A \qiq M_{ij} = \sk\sl N_{\c{k\ell}}\,\A_{\c{k\ell}ij} \\ \|A\|_F^2 &= A:A \qiq \|A\|_F^2 = \sk\sl A_{\c{k\ell}}\,A_{\c{k\ell}} \\ }$$ The product of an arbitrary matrix $M$ with the fourth-order isotropic tensors yields $$\eqalign{ \E:M &= M:\E &= M \\ \F:M &= M:\F &= M^T \\ \G:M &= M:\G &= I\,\trace{M} \\ }$$ Thus $\E$ is sometimes called the identity tensor since it acts as the identity here.
Note that the $\I$ in your post actually equals $\G$, which is one of the isotropic tensors but not the identity tensor (although it does yield a scalar multiple of the identity matrix).
Some authors refer to the following $\S$ tensor as the identity tensor $$\eqalign{ \S &= \tfrac 12\LR{\E+\F} \\ M:\S &= \S:M = \frac{M+M^T}{2} \;\equiv\; \sym{M} \\ }$$ but as you can see, it acts as an identity only if the matrix $M$ is symmetric.
Another useful property of the isotropic tensors is their ability to rearrange matrix-vector products. For example $$\eqalign{ P\cdot X\cdot Q &= \LR{P\cdot\E\cdot Q^T}:X \\ P\cdot X\cdot q &= \LR{P\cdot\E\cdot q}:\F:X &= \LR{P\cdot\G\cdot q}:X \\ p\cdot X\cdot Q &= \LR{p\cdot\E\cdot Q^T}:X &= X:\LR{p\cdot\G\cdot Q} \\ p\cdot X\cdot q &= \LR{p\cdot\G\cdot q}:X &= \LR{p\cdot q^T}:X \\ p\cdot x^T\cdot Q &= \LR{p\cdot\G\cdot Q^T}\cdot x \\ P\cdot x\cdot q^T &= \LR{P\cdot\E\cdot q}\cdot x \\ }$$ where $(x,p,q)$ are vectors and $(X,P,Q)$ are matrices.