Question

explain tensor. prove that the sum of two tensor is a tensor. show that by contraction , the rank of tensor is reduced by two.

Answer 1

Explain tensor. prove that the sum of two tensor is a tensor. Show that by contraction, the rank of tensor is reduced by two.

Solution

$$A _ {i _ {n + 1} \dots i _ {n + m}} ^ {i _ {1} \dots i _ {n}}, B _ {i _ {n + 1} \dots i _ {n + m}} ^ {i _ {1} \dots i _ {n}} \text{ is tensors of (n,m)-type.}$$

According to the transformation law of tensors we have

$$\begin{array}{l} \tilde {A} _ {i _ {n + 1} \dots i _ {n + m}} ^ {i _ {1} \dots i _ {n}} = (R) ^ {- 1 ^ {i _ {1}}} \dots (R) ^ {- 1 ^ {i _ {n}}} R _ {i _ {n + 1}} ^ {j _ {n + 1}} \dots R _ {i _ {n + m}} ^ {j _ {n + m}} A _ {j _ {n + 1} \dots j _ {n + m}} ^ {j _ {1} \dots j _ {n}} \\ \tilde {B} _ {i _ {n + 1} \dots i _ {n + m}} ^ {i _ {1} \dots i _ {n}} = (R) ^ {- 1 ^ {i _ {1}}} \dots (R) ^ {- 1 ^ {i _ {n}}} R _ {i _ {n + 1}} ^ {j _ {n + 1}} \dots R _ {i _ {n + m}} ^ {j _ {n + m}} B _ {j _ {n + 1} \dots j _ {n + m}} ^ {j _ {1} \dots j _ {n}} \\ \end{array}$$

Where $R_{i}^{j}$ is a matrix. The components, $\nu^{i}$, of a regular (or column) vector, $\mathbf{v}$, transform with the inverse of the matrix $R$,

\widehat {\boldsymbol {v}} ^ {i} = \left(R\right) ^ {- 1} _ {j} ^ {i} \boldsymbol {v} ^ {j}

where the hat denotes the components in the new basis. While the components, $w_{n}$ of a covector (or row vector), $\mathbf{w}$ transform with the matrix R itself,

$$\begin{array}{l} \widehat {\boldsymbol {w}} _ {i} = R _ {i} ^ {j} \boldsymbol {w} _ {j} \\ \Rightarrow \\ \alpha A _ {j _ {1} \dots j _ {k}} ^ {i _ {1} \dots i _ {n}} + \beta B _ {j _ {1} \dots j _ {k}} ^ {i _ {1} \dots i _ {n}} = (R) ^ {- 1 ^ {i _ {1}}} \dots (R) ^ {- 1 ^ {i _ {n}}} R _ {j _ {n + 1}} ^ {j _ {n + 1}} \dots R _ {j _ {n + m}} ^ {j _ {n + m}} (\alpha A _ {j _ {n + 1} \dots j _ {n + m}} ^ {j _ {1} \dots j _ {n}} + \beta B _ {j _ {n + 1} \dots j _ {n + m}} ^ {j _ {1} \dots j _ {n}}) \\ \end{array}$$

$\alpha, \beta$ is constant

From hence $\alpha A_{j_{n + 1}\dots j_{n + m}}^{j_1\dots j_n} + \beta B_{j_{n + 1}\dots j_{n + m}}^{j_1\dots j_n}$ transformed as tensor.

From hence $\alpha A_{j_{n + 1}\dots j_{n + m}}^{j_1\dots j_n} + \beta B_{j_{n + 1}\dots j_{n + m}}^{j_1\dots j_n}$ is tensor

Contraction is $A_{i_1\dots i_{n + m}}^{i_1\dots i_n}$

$$\begin{array}{l} \tilde {A} _ {i _ {1} \dots i _ {n + m}} ^ {i _ {1} \dots i _ {n}} = (R) ^ {- 1 ^ {i _ {1}}} \dots (R) ^ {- 1 ^ {i _ {n}}} R _ {i _ {1}} ^ {j _ {1}} \dots R _ {i _ {n + m}} ^ {j _ {n + m}} A _ {j _ {1} \dots j _ {n + m}} ^ {j _ {1} \dots j _ {n}} \Rightarrow \\ \tilde {A} _ {i _ {1} \dots i _ {n + m}} ^ {i _ {1} \dots i _ {n}} = (R) ^ {- 1 ^ {i _ {2}}} \dots (R) ^ {- 1 ^ {i _ {n}}} R _ {j _ {n}} ^ {j _ {n + 2}} \dots R _ {i _ {n + m}} ^ {j _ {n + m}} A _ {j _ {1} \dots j _ {n + m}} ^ {j _ {1} \dots j _ {n}} \\ \end{array}$$

Because $(R)^{-1^{j_1}} * R_{j_1}^{j_1} = E$ is identity matrix.

From hence $A_{j_1 \cdots j_{n-m}}^{j_1 \cdots j_n}$ transformed as tensor $(n-1, m-1)$ rank.

From hence $A_{j_1 \cdots j_{n-m}}^{j_1 \cdots j_n}$ is $(n-1, m-1)$ rank.