Evaluating contractions of a tensor product Consider $T = \delta \otimes \gamma$ where $\delta$ is the $(1,1)$ Kronecker delta tensor and $\gamma \in T_p^*(M)$, the co-tangent space over some manifold $M$.  Evaluate all possible contractions of $T$.
Attempt:
$\gamma$ is therefore a $(0,1)$ tensor and the tensor product with $(1,1)$ yields a $(1,2)$ tensor.  The components of $T$ are therefore $T^a_{\,\,\,\,\,bc}$ which gives rise to the two possible contractions $T^a_{\,\,\,\,\,ab}$ or $T^a_{\,\,\,\,\,ba}$.  Do I need to include any more detail?  
 A: Here is a concrete index version of my answer, given in the comments to the question.
Let $\{e_i \}$ be a local frame around some point in manifold $M$, and $\{f^j \}$ is the corresponding dual local co-frame, that is $f^{i}(e_{j}) = \delta^{i}{}_{j}$, where  $\delta^{i}{}_{i} = 1$ and $\delta^{i}{}_{j} = 0$ if $i\neq j$, for all $i,j = 1, \dots, n$.
The Kronecker symbol $\delta^i{}_j$ represents the local components of the identity operator $\delta \colon T M \to T M$ in this frame/co-frame pair, that is locally $\delta = \delta^i{}_j e_i \otimes f^j$. The Einstein summation convention is used in such expressions.
The one form $\gamma$ in local components is given by $\gamma = \gamma_k f^k$.
Tensor $T = \delta \otimes \gamma$ is then represented locally as $T = \delta^i{}_j \gamma_k e_i \otimes f^j \otimes f^k$, that is $T^{i}{}_{j k} = \delta^i{}_j \gamma_k$ for $i,j,k = 1, \dots, n$.
In our local frame we have a frame $\{e_i \otimes f^j \otimes f^k\}$ for the tensor bundle $T M \otimes T^* M \otimes T^* M$, and we see that the $1$-st factor is a vector, and the $2$-nd and $3$-rd factors are co-vectors. 
A contraction operator $C(p,q)$ is a member of a whole family of operators on tensor spaces (bundles), which can be defined on elements of the frame as taking the $p$-th factor and substituting it into the $q$-th factor (and this operation must make sense!), and then extending this action to all tensors by the (multi-) linearity.
In our case we have only two possible contraction operators, namely $C(1,2)$ and $C(1,3)$.
Applying them to our tensor, we compute:
$$
C(1,2)(T) = \delta^i{}_j \gamma_k f^j(e_i ) f^k = \delta^i{}_j \gamma_k \delta^j{}_i f^k = n \, \gamma
$$
where we have used that $\delta^i{}_j  \delta^j{}_i= \sum_{i=1}^n \sum_{j = 1}^{n} \delta^i{}_j  \delta^j{}_i = \sum_{i=1}^n \delta^i{}_i = \sum_{i=1}^n 1 = n$.
The case $C(1,3)(T)$ is similar, but simpler.
