Constructing a basis for the space of (k,l) tensors

Question

The book I'm reading on general relativity introduces tensors and define a tensor basis like this:

The space of all tensors of a fixed type (k,l) forms a vector space. To construct a basis for the space of (k,l) tensors is necessary to to introduce the tensor product. If T is a (k,l) tensors and S is a (m,n) tensor we define a (k+m,l+m) tensor $T\otimes S$

\begin{equation}T\otimes S (\omega^{(1)},\ldots, \omega^{(k+m)}, V^{(1)},V^{(l+n)}) = T (\omega^{(1)},\ldots, \omega^{(k)}, V^{(1)},V^{(l)}) \\ \times S (\omega^{(k+1)},\ldots, \omega^{(k+m)}, V^{(l+1)},V^{(l+n)})\end{equation}

It is now straightforward to construct a basis for the space of all (k,l) tensors, by taking the tensor products of basis vectors and dual vectors: this basis will consist of all tensors of the form:

\begin{equation}e_{(\mu_1)}\otimes \ldots \otimes e_{(\mu_k)} \otimes \theta^{(\nu_1)}\otimes \ldots \otimes \theta^{(\nu_l)}\end{equation}

An arbitrary tensor can be written as:

$T = T^{\mu_1 \ldots\mu_k}_{\nu_1\ldots\nu_k}e_{(\mu_1)}\otimes \ldots \otimes e_{(\mu_k)} \otimes \theta^{(\nu_1)}\otimes \ldots \otimes \theta^{(\nu_l)}$

It is not straightforward for me to understand why it is defining the basis like that. Could you please show me an example of spacetime tensor expressed by its basis tensors like the metric tensor? thank you

Answer 1

If $(e_\mu)$ is a basis and $(\theta^\nu)$ is the dual basis, the components of a tensor $T$ with respect to these bases are defined to be $$T_{\nu_1\cdots \nu_\ell}^{\mu_1\cdots \mu_k} = T(\theta^{\mu_1},\ldots, \theta^{\mu_k},e_{\nu_1},\ldots, e_{\nu_\ell}).$$By multilinearity of $T$, we have that given vectors $v_1,\ldots, v_\ell$ and covectors $\xi^1,\ldots, \xi^k$, the following holds: $$\begin{align} T(\xi^1,\ldots, \xi^k,v_1,\ldots,v_\ell) &= T(\xi^1_{\mu_1}\theta^{\mu_1},\ldots, \xi^k_{\mu_k}\theta^{\mu_k}, v_1^{\nu_1}e_{\nu_1},\ldots, v_\ell^{\nu_\ell}e_{\nu_\ell}) \\ &= \xi^1_{\mu_1}\cdots \xi^k_{\mu_k}v_1^{\nu_1}\cdots v_\ell^{\nu_\ell} T(\theta^{\mu_1},\ldots, \theta^{\mu_k},e_{\nu_1},\ldots, e_{\nu_\ell}) \\ &= T_{\nu_1\cdots \nu_\ell}^{\mu_1\cdots \mu_k}\xi^1_{\mu_1}\cdots \xi^k_{\mu_k}v_1^{\nu_1}\cdots v_\ell^{\nu_\ell}\\ &= T_{\nu_1\cdots \nu_\ell}^{\mu_1\cdots \mu_k} \xi^1(e_{\mu_1})\cdots \xi^k(e_{\mu_k})\cdots\theta^{\nu_1}(v_1)\cdots \theta^{\nu_\ell}(v_\ell) \\&= T_{\nu_1\cdots \nu_\ell}^{\mu_1\cdots \mu_k} (e_{\mu_1}\otimes \cdots \otimes e_{\mu_k}\otimes \theta^{\nu_1}\otimes \cdots \otimes \theta^{\nu_\ell})(\xi^1,\ldots, \xi^k,v_1,\ldots, v_\ell), \end{align}$$which shows that $$T = T_{\nu_1\cdots \nu_\ell}^{\mu_1\cdots \mu_k}\,e_{\mu_1}\otimes \cdots \otimes e_{\mu_k}\otimes \theta^{\nu_1}\otimes \cdots \otimes \theta^{\nu_\ell}.$$That is to say, we have proved that $\{e_{\mu_1}\otimes \cdots \otimes e_{\mu_k}\otimes \theta^{\nu_1}\otimes \cdots \otimes \theta^{\nu_\ell} \}$ generates the space of $(k,\ell)$-tensors. To prove that it is linearly independent (and hence a basis), assume that $$a_{\nu_1\cdots \nu_\ell}^{\mu_1\cdots \mu_k} \,e_{\mu_1}\otimes \cdots \otimes e_{\mu_k}\otimes \theta^{\nu_1}\otimes \cdots \otimes \theta^{\nu_\ell} = 0$$for some coefficients $a_{\nu_1\cdots \nu_\ell}^{\mu_1\cdots \mu_k}$. Our goal is to show that these coefficients vanish. Well, evaluate both sides of the above expression in $(\theta^{\alpha_1},\ldots,\theta^{\alpha_k},e_{\beta_1},\cdots, e_{\beta_\ell})$ to get $$0 = a_{\nu_1\cdots \nu_\ell}^{\mu_1\cdots \mu_k}\delta_{\mu_1}^{\alpha_1}\cdots \delta_{\mu_k}^{\alpha_k}\delta_{\beta_1}^{\nu_1}\cdots \delta^{\nu_\ell}_{\beta_\ell} = a_{\beta_1\cdots \beta_\ell}^{\alpha_1\cdots \alpha_k},$$and we are done by the arbitrariety of the indices $\alpha$ and $\beta$.