In this post I wanted to get a quick and dirty intuition of the mechanics of a tensor acting on two vectors:
Let $\beta \in V^*$ with coordinates in standard Cartisian basis $\beta=\color{blue}{\begin{bmatrix}\sqrt{\pi} & \sqrt[3]{\pi} &\sqrt[5]{\pi} \end{bmatrix}}$ and $\gamma\in V^*$ with coordinates $\gamma=\color{red}{\begin{bmatrix}\frac{1}{3} &\frac{1}{5} &\frac{1}{7} \end{bmatrix}}$. Now if we apply the tensor product $T_{\mu\nu}\,e^\mu\otimes\,e^\nu=
\beta\otimes \gamma$ on the vectors
$$v=\color{magenta}{\begin{bmatrix}1\\7\\5\end{bmatrix}}, w = \color{orange}{\begin{bmatrix}2\\0\\3\end{bmatrix}}$$
we get
$$(\beta \otimes \gamma)[v,w]=\langle \color{blue}\beta,\color{magenta}v \rangle \times \langle \color{red}\gamma,\color{orange}w\rangle.$$
So two dot products multiplied.
If instead we had a tensor
$$T_{\mu\nu}{}^{\rho}\,e^\mu\otimes\,e^\nu\otimes e_\rho= \beta\otimes \gamma\otimes v ,$$
and we contracted $\beta$ and $v,$ would it be correct that we'd end up with a tensor $$T_{\nu}\,e^\nu
=\langle \color{blue}\beta,\color{magenta}v \rangle \,\gamma$$
?
In other words, the contraction contributes to a scalar that comes up front as a scalar multiplying à tensor reduced in rank by $2$ after leaving out the two contracted elements (in this case, just a covector)?
As an amateur learner it is impossible not to revisit topics every few years. So I would not have asked this question today, and for whomever may read this, I want to point out as a NB that:\
-
Tensor contraction is properly understood as an operation within a tensor not between a tensor and another tensor (e.g. vector or covector).
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Contracting a tensor is the equivalent of obtaining the trace of a matrix – it says something important about the matrix, but it reduces the matrix to a scalar, and hence, there is a loss of information. In the case of any other tensor, e.g. $T^{\mu\lambda}{}_{\nu\gamma}$, one of the contravariant indices, e.g. $\lambda,$ can be contracted with a covariant index such as $\gamma,$ provided they are of the same size by turning them into dummy indices $T^{\mu i}{}_{\nu i}$ and summing over (in 3 dimensional space):
$$\left(\sum_{i=1}^3 A^\lambda A_\gamma\delta^\lambda_\gamma \right) T^{\mu}{}_{\nu}$$
or
$$P^{\mu }{}_{\nu } = T^{\mu 1}{}_{\nu 1} + T^{\mu 2}{}_{\nu 2} + T^{\mu 3}{}_{\nu 3}$$
Best Answer
Yes, a contraction over $\beta$ and $v$ would result in $\langle \beta, v \rangle \gamma$: $$ \delta_\rho^\mu T_{\mu\nu}{}^{\rho} = \delta_\rho^\mu \beta_\mu \gamma_\nu v^\rho = \beta_\mu \gamma_\nu v^\mu = \langle \beta, v \rangle \gamma_\nu =: T_\nu . $$