# How to prove an expression to be a tensor?

How to prove that the expression $$\varphi_{,ij}:=\frac{\partial^2\varphi}{\partial x_i\partial x_j}=\nabla\nabla\varphi$$ is a tensor of second order where $$\varphi$$ is a scalar? Furthermore, how to prove that $$a\times b:=a_i b_j\varepsilon_{ijk}e_k$$ is a vector?

We can either prove it by definition or use the so-called "tensor recognition theorem" claiming that if $$p_{i_1i_2\cdots i_mj_1j_2\cdots j_n}q_{j_1\cdots j_n} = r_{i_1\cdots i_m}$$, then $$p$$ must be a tensor of order $$m+n$$, where $$q_{j_1\cdots j_n}$$ is a tensor of order $$n$$ and $$r_{i_1\cdots i_m}$$ a tensor of order $$m$$.

$$1$$) Suppose that $$O X_1 X_2 X_3$$ is the coordinate system corresponding to the given basis $$e_1, e_2, e_3$$ on $$E_3$$. The gradient of a scalar function $$\varphi (X_1,X_2,X_3)$$, using this coordinate system, is defined by $$\nabla \phi(X_1,X_2,X_3) = \frac{\partial \varphi}{\partial X_i} e_i.$$

Suppose that $$\tilde{A} = \{ \tilde{e}_1,\tilde{e}_2,\tilde{e}_3 \}$$ is another orthonormal basis with corresponding cartesian coordinates $$O \tilde{X_1} \tilde{X_2} \tilde{X_3}$$ given by $$\tilde{X}_i = q_{ij} X_j$$, where $$Q = (q_{ij}) \in SO(3)$$.

By the chain rule, $$\frac{\partial \tilde{\varphi}}{\partial \tilde{X}_i} = \frac{\partial \varphi}{\partial X_k} \frac{\partial X_k}{\partial \tilde{X}_i}.$$

Using $$\tilde{X}_i = q_{ij} X_j$$, we have $$\frac{\partial X_k}{\partial \tilde{X}_i} = q_{ik},$$ and hence $$\frac{\partial \tilde{\varphi}}{\partial \tilde{X}_i} = q_{ik} \frac{\partial \varphi}{\partial X_k}.$$

That is, by definition, $$\nabla \varphi$$ is a cartesian tensor of order $$1$$. Doing this again, but with $$\nabla \varphi$$ in place of $$\varphi$$, shows that $${\nabla}^{2} \varphi$$ is a cartesian tensor of order $$2$$.

$$2$$) $$a$$ and $$b$$ are vectors so $$a_i$$ and $$b_j$$ are components of cartesian tensors of order $$1$$, and the alternating tensor $$\varepsilon = \varepsilon_{klm}$$ is a cartesian tensor of order $$3$$.

Taking their product we get that $$a_i b_j \varepsilon_{klm}$$ are the components of a cartesian tensor of order $$5$$.

Contracting indices (that is, setting two indices equal and thus effecting a sum) gives that $$a_i b_j \varepsilon_{ilm}$$ are the components of a cartesian tensor of order $$4$$, and contracting again gives that $$a_i b_j \varepsilon_{ijm}$$ are the components of a cartesian tensor of order $$3$$.

Hence, $$a \times b$$ is a cartesian tensor of order $$3$$.