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Let $T$ be a $(1,1)$ tensor over a vector space $V$. Let $\left\{e_a\right\}$ be a basis for $V$ and $\left\{f^a\right \}$ be its dual basis. Show that $T$ is equivalent to a linear map $V^* \rightarrow V^*$. Similarly, show that $T$ is also equivalent to a linear map $V \rightarrow V$.

Attempt: $T$ being a $(1,1)$ tensor means that it is a linear map $V^* \times V \rightarrow \mathbb{R}$ or explicitly, I think I can write $T : (\lambda, X) \mapsto \lambda_a X^aT^i_{\,\,j} \in \mathbb{R}$ where $T^i_{\,\,j}, \lambda_a$ and $X^a$ are coefficients of the tensor, co-vector and vector respectively in basis $\left\{f^a, e_a\right\}$ .

Now the book goes onto write the following:

Given a $(1, 1)$ tensor $T$ and $\lambda \in V^*$, we can form the $(0, 1)$ tensor $T(\lambda, \cdot)$, i.e. $T(\lambda, \cdot) \in V^*$. Thus $T$ defines a linear map $V^* \rightarrow V^*$, $λ \mapsto T(\lambda, \cdot).$ I don't really understand this paragraph. How can such a tensor be formed? It looks like some sort of contraction, however contraction of a $(1,1)$ tensor would only produce $(0,0)$ tensor as far as I understand. (Contraction of $T^i_{\,\,j} \rightarrow T^i_{\,\,i}$). Similarly, why is $T(\lambda, \cdot)$ even a $(0,1)$ tensor? As far as I understand, the notation $(0,1)$ means the map feeds on zero covector arguments and one vector argument which is not what $T(\lambda, \cdot)$ suggests given that $\lambda \in V^*$.

Many thanks!

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1 Answer 1

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Let $V$ be a real, finite-dimensional vector space $V$ and $$T: V^* \otimes V \to \mathbb{R}$$ a $(1, 1)$-tensor.

Now, given any $\lambda \in V^*$, we denote by $$T(\lambda, \,\cdot\,)$$ the map $V \to \mathbb{R}$ defined by $$X \mapsto T(\lambda, X).$$ Since $T$ is multilinear, the map $T(\lambda, \,\cdot\,)$ is linear, but linear maps $V \to \mathbb{R}$ are precisely the elements of $V^*$. Thus, the map $$\lambda \mapsto T(\lambda, \cdot)$$ is a map $V^* \to V^*$, and again because $T$ is multilinear this map is linear.

In the bases $(e_a)$ and $(f^a)$ we can decompose (using the Einstein summation convention) $\lambda = \lambda_a f^a$ and $T = T^a{}_b e_a f^b$. Now, for all $X = X^a e_a$, we have $$T(\lambda, \,\cdot\,)(X) = T(\lambda, X) = T^a{}_b \lambda_a X^b.$$ On the other hand (using slightly peculiar notation), we can decompose $T(\lambda, \,\cdot\,)(X) = T(\lambda, \,\cdot\,)_b X^b$. So, $T^a{}_b \lambda_a X^b = T(\lambda, \,\cdot\,)_b X^b$ for all $X$ and hence the components of $T(\eta, \,\cdot\,)$ w.r.t. the dual basis $f^b$ are: $$T(\lambda, \,\cdot\,)_b = \lambda_a T^a{}_b.$$

In particular, this shows that the map $\lambda \mapsto T(\lambda, \,\cdot\,)$ can be described by a contraction, not of $T$ alone but as the map that sends $\lambda$ to the only possible contraction of $T \otimes \lambda$.

On the other hand, the sum $$\lambda_a T^a{}_b$$ is precisely the $b$th component of the product $[\lambda] [T]$ of the matrix representations $[\lambda]$ and $[T]$ resp. of $\lambda$ and $T$ w.r.t. the basis $(e_a)$. In other words, w.r.t. a basis the map $V^* \to V^*$ is simply matrix multiplication (on the right) of a row vector by $[T]$.

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