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 Post subject: Tensors - Part 2
PostPosted: Sun Dec 25, 2016 6:25 pm 
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Definition 1: Let $V$ be a finite-dimensional $\mathbb{R}$-vector space and let $k$ be a non-negative integer. A covariant $k$-tensor is a multilinear function $T \ \colon V \times \dots \times V \to \mathbb{R} $.

Definition 2: A covariant $k$-tensor is called alternating if its value changes sign by interchanging any pair of its arguments:
\[ T(X_{1}, \dots, X_{i}, \dots, X_{j}, \dots, X_{k}) = - T(X_{1}, \dots, X_{j}, \dots, X_{i}, \dots, X_{k}) \]whenever $ 1 \leq i < j \leq k $.



Show that the following are equivalent for a covariant $k$-tensor $T$:
  1. $T$ is alternating.
  2. For any vectors $X_{1}, \dots, X_{k} \in V $ and any permutation $ \sigma \in S_{k} $, it holds that
    \[ T( X_{\sigma(1)}, \dots, X_{\sigma(k)}) = \text{sgn}(\sigma) T(X_{1}, \dots, X_{k}) \]
  3. $T$ gives zero whenever two of its arguments are equal, that is
    \[ T(X_{1}, \dots, Y, \dots, Y, \dots X_{k}) = 0 \]
  4. $ T(X_{1}, \dots, X_{k}) = 0 $ whenever the vectors $ X_{1}, \dots, X_{k} \in V $ are linearly dependent.
  5. The components $T_{i_{1} \dots i_{k}} = T(E_{i_{1}}, \dots, E_{i_{k}})$ of $T$ with respect to any basis $ \left\{ E_{i} \right\} $ of $V$ change sign whenever two indices are interchanged.


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