# How is it that two covariant vectors can be multiplied without one being transposed?

This reading, without much explanation, claims that given two covariant vectors,

$$\tilde x = \begin{pmatrix} x_1 & x_2 & x_3 \end{pmatrix}$$ $$\tilde y = \begin{pmatrix} y_1 & y_2 & y_3 \end{pmatrix}$$

one has the product

$$x_\mu y_\nu = \begin{pmatrix} x_1\cdot y_1 & x_1\cdot y_2 & x_1\cdot y_3 \\ x_2\cdot y_1 & x_2\cdot y_2 & x_2\cdot y_3 \\ x_3\cdot y_1 & x_3\cdot y_2 & x_3\cdot y_3 \\ \end{pmatrix}$$

Having incredibly little background knowledge on this subject, this makes no sense to me. How can $1\times3$ array times a $1\times3$ array yield a $3\times3$ array? Furthermore, this looks just like the product of a covector and a contravector to me.

Could someone shed some light on what’s going on here?

Its achieved by forming tensor product of two covector. The tensor product can be thought as generalization of outer product of vectors. Suppose we have a vector space $V$ with its dual space $V^*$, the elements $x \in V^*$ is a linear map $$x : V \rightarrow \mathbb{R}.$$
If we have two covectors $\tilde{x},\tilde{y} \in V^*$ where in terms of basis we write them as $\tilde{x} = x_{\mu} \varepsilon^{\mu}$ and $\tilde{y} = y_{\mu}\varepsilon^{\mu}$ (with $\varepsilon^{\mu}$ is the dual basis of $V^*$), their tensor product $\tilde{x} \otimes \tilde{y}$ is $(0,2)$-covariant tensor $$\tilde{x} \otimes \tilde{y} : V \times V \rightarrow \mathbb{R}$$ defined as \begin{align} (\tilde{x} \otimes \tilde{y}) (v,w) &= \tilde{x}(v) \tilde{y}(w) \\ &= x_{\mu} \varepsilon^{\mu}(v) \, y_{\nu} \varepsilon^{\nu}(w) \\ &= x_{\mu} y_{\nu} \, \varepsilon^{\mu}(v)\varepsilon^{\nu}(w) \qquad \text{for all} \, v,w \in V \end{align} so in components, we write as $\tilde{x} \otimes \tilde{y} = x_{\mu}y_{\nu} \, \varepsilon^{\mu} \otimes \varepsilon^{\nu}$. We can easily verify that $\tilde{x} \otimes \tilde{y}$ is a multilinear map (i.e $\tilde{x} \otimes \tilde{y}$ is a tensor), and the collection $$\{ \varepsilon^{\mu} \otimes \varepsilon^{\nu} \mid \mu,\nu = 1,2,3 \}$$ is a basis for the tensor space $(0,2)$. Your $3\times 3$ array above is the components of this tensor. The author in your article above define tensor as an object that transform properly. You can check that if $x_{\mu}$ and $y_{\nu}$ transform as $x'_{\alpha} = B_{\alpha}^{\mu}x_{\mu}$ and $y'_{\beta} = B_{\beta}^{\nu} y_{\nu}$ and if you define a new object $t_{\mu\nu} = x_{\mu}y_{\nu}$, then this object transform properly as $$t'_{\alpha \beta} = x'_{\alpha}y'_{\beta} = B_{\alpha}^{\mu}x_{\mu} B_{\beta}^{\nu} y_{\nu} = B_{\alpha}^{\mu}B_{\beta}^{\nu} x_{\mu} y_{\nu} = B_{\alpha}^{\mu}B_{\beta}^{\nu} t_{\mu\nu} .$$ We can form any tensor product of any two (or more) tensor (of any type) by the same way (look for example P. Szekeres or Lee's Smooth Manifold for more details).
However, based on the approach of the text above, you dont really need to know what is tensor product to know that $t_{\mu\nu}=x_{\mu}y_{\nu}$ is a tensor. As long as the object transform properly (even if the construction is look strange to you as $t_{\mu\nu}$ above), you can be sure that its a tensor. This is the kind of approach that the author follow.