Tensor contraction criteria On Carroll's Spacetime and Geometry page 25, the book introduces tensor contraction, which proceeds by summing over one upper and one lower index:
$$S^{\mu \rho}_{\ \ \ \ \  \sigma}=T^{\mu \nu \rho}_{\ \ \ \ \ \ \ \   \sigma \nu} \tag{1.70}$$
The book said the result is a well-defined tensor and

By well-defined tensor we mean either "transforming according to the tensor transformation law," or "defining a unique multilinear map from a set of vectors and dual vectors to the real numbers"

I tried the first criteria:
$$T^{\mu^{\prime} \nu^{\prime} \rho^{\prime}}_{\ \ \ \ \ \ \ \ \ \ \   \sigma^{\prime} \nu^{\prime}}=\Lambda^{\mu^{\prime}}_{\ \ \  \mu}\Lambda^{\nu^{\prime}}_{\ \ \  \nu}\Lambda^{\rho^{\prime}}_{\ \ \  \rho}\Lambda^{\sigma}_{\ \ \  \sigma^{\prime}}\Lambda^{\nu}_{\ \ \  \nu^{\prime}}T^{\mu \nu \rho}_{\ \ \ \ \ \ \ \   \sigma \nu} \tag{A}$$
I expect that the Lorentz transformation matrices can exchange with each other and we can use the property
$$ \Lambda^{\nu^{\prime}}_{\ \ \  \nu}\Lambda^{\nu}_{\ \ \  \nu^{\prime}}=\text{Number (trace of )} \delta^{\nu^{\prime}}_{\nu^{\prime}} \tag{B}$$
But I am troubled for the following sentence in the book

Note also that the order of the indices matters, so that you can get different tensors by contracting in different ways; thus $$T^{\mu \nu \rho}_{\ \ \ \ \ \ \ \   \sigma \nu}\neq T^{\mu \rho \nu }_{\ \ \ \ \ \ \ \   \sigma \nu} \tag{1.71}$$

I can understand this order really matters from tensor's definition. But how to understand this order in contraction? Since we only have one form of $S^{\mu \rho}_{\ \ \ \ \  \sigma}$, can this represent different orders? This also implies my exchange of $\Lambda$ matrices is wrong!
 A: 
I tried the first criteria:
$$T^{\mu^{\prime} \nu^{\prime} \rho^{\prime}}_{\ \ \ \ \ \ \ \ \ \ \   \sigma^{\prime} \nu^{\prime}}=\Lambda^{\mu^{\prime}}_{\ \ \  \mu}\Lambda^{\nu^{\prime}}_{\ \ \  \nu}\Lambda^{\rho^{\prime}}_{\ \ \  \rho}\Lambda^{\sigma}_{\ \ \  \sigma^{\prime}}\Lambda^{\nu}_{\ \ \  \nu^{\prime}}T^{\mu \nu \rho}_{\ \ \ \ \ \ \ \   \sigma \nu} \tag{A}$$

What you wrote about doesn't make sense for at least two reasons. First, the tensor on the LHS should be denoted with a different symbol, such as $\tilde T$ to indicate it is transformed. Second, you probably shouldn't use four different $\nu$ indices on the RHS, since it is not clear which are contracted with which.
What you want to consider is:
$$
{\tilde T}^{\mu^{\prime} \nu^{\prime} \rho^{\prime}}_{\ \ \ \ \ \ \ \ \ \ \   \sigma^{\prime} \tau^{\prime}}=\Lambda^{\mu^{\prime}}_{\ \ \  \mu}\Lambda^{\nu^{\prime}}_{\ \ \  \nu}\Lambda^{\rho^{\prime}}_{\ \ \  \rho}\Lambda^{\sigma}_{\ \ \  \sigma^{\prime}}\Lambda^{\tau}_{\ \ \  \tau^{\prime}}T^{\mu \nu \rho}_{\ \ \ \ \ \ \ \   \sigma \tau}
$$
and therefore:
$$
{\tilde T}^{\mu^{\prime} \nu^{\prime} \rho^{\prime}}_{\ \ \ \ \ \ \ \ \ \ \   \sigma^{\prime} \nu^{\prime}}=\Lambda^{\mu^{\prime}}_{\ \ \  \mu}\Lambda^{\nu^{\prime}}_{\ \ \  \nu}\Lambda^{\rho^{\prime}}_{\ \ \  \rho}\Lambda^{\sigma}_{\ \ \  \sigma^{\prime}}\Lambda^{\tau}_{\ \ \  \nu^{\prime}}T^{\mu \nu \rho}_{\ \ \ \ \ \ \ \   \sigma \tau}\;,
$$
where there is an implicit sum over $\nu'$ on both the LHS and the RHS. (N.b., a different letter could be used for the dummy variable if desired.).
Next, the (implicit) sum over $\nu'$ on the RHS is used to simplify:
$$
\Lambda^{\nu^{\prime}}_{\ \ \  \nu}\Lambda^{\tau}_{\ \ \  \nu^{\prime}} = \delta_{\nu}^\tau\;,
$$
Now the sum over $\tau$ is trivial and we see that:
$$
{\tilde T}^{\mu^{\prime} \nu^{\prime} \rho^{\prime}}_{\ \ \ \ \ \ \ \ \ \ \   \sigma^{\prime} \nu^{\prime}}=\Lambda^{\mu^{\prime}}_{\ \ \  \mu}\Lambda^{\rho^{\prime}}_{\ \ \  \rho}\Lambda^{\sigma}_{\ \ \  \sigma^{\prime}}T^{\mu \nu \rho}_{\ \ \ \ \ \ \ \   \sigma \nu}\;.
$$
In other words
$$
{T}^{\mu \alpha \rho}_{ \ \ \ \ \ \ \ \   \sigma \alpha}\;,
$$
where there is an implicit sum on the dummy variable alpha,
transforms like a tensor with two upper indices and one lower index.
A: You can contract every pair of indices, and if one is covariant and one is contravariant this operation runs quite smoothly in index notation, without the appearance of metric tensor.
When you perform the contraction of two indices you should declare which pair of indices is involved, since the result changes in general (if no symmetry between the indices exists) if you change the pair of indices you're contracting.
As an example, on a third-order tensor $\mathbb{A} = A^{ij}_{\ \ k} \mathbf{b}_i \otimes \mathbf{b}_j \otimes \mathbf{b}^k$,

*

*the contraction of the first and third indices reads $C^1_3  (\mathbb{A}) = A^{ij}_{\ \ i} \mathbf{b}_j$, i.e. the $j^{th}$ component of the resulting vector is $A^{ij}_{\ \ i}$, with the sum over $i$;


*the contraction of the second and third indices reads $C^2_3 (\mathbb{A}) = A^{ji}_{\ \ i} \mathbf{b}_j$, i.e. the $j^{th}$ component of the resulting vector is $A^{ji}_{\ \ i}$, with the sum over $i$; if the tensor $\mathbb{A}$ has no symmetry in the first pair of indices $A^{ji}_{\ \ i} \ne A^{ij}_{\ \ i}$ and thus the components of the resulting vectors expressed in the same basis are different, and thus the result is different as well.
A: The tensor can indeed be non symmetric, as well as the Lorentz transformation.
This only mean that you can't change the order of indices in each tensor but you can still change the order of the different
components of tensors keeping the includes in place.
The calculation you expect still work as you can exchange the order of each pair of components of the Lorentz transformation and use the id identity you mentioned
