I'm a beginner in tensors, and all my concepts of tensors are from the book: "Munkres-Analysis on Manifolds"
Consider the $2$ tensor field $\omega$ on $\mathbb{R}^3$ . Then, from my book, it means that for each vector $x\in \mathbb{R}^3$, $\omega(x)$ represents a $2$ tensor on the tangent space of $x$ denoted as $\mathscr{T}_{x}(\mathbb{R}^3)$. That is, if use the notation $(x;v)$ to represent the vectors in the tangent space $\mathscr{T}_{x}(\mathbb{R}^3)$, we have $$\begin{align} \omega(x):((x;v_{1}),(x;v_{2})) \to r \end{align}$$ where $r$ is a number on $\mathbb{R}$
Now I let the vectors $v_{1},v_{2}$ depend on $x$ , so $(x;v_{1}), \ (x;v_{2})$ become 2 vector fields $$\begin{align} F_{1}(x)=(x; v_{1}(x)) \\ \\ F_{2}(x)=(x;v_{2}(x)) \end{align}$$ on $\mathbb{R}^3$ where each of then is in the tangent space of $x$, and $$\begin{align} v_{1}:& \mathbb{R}^3 \to \mathbb{R}^3 \\ :& x=(x^1,x^2,x^3) \to (v_{1}^1(x),v_{1}^2(x),v_{1}^3(x))=v_{1}(x) \end{align}$$ So does $v_{2}$.
Then I define a operator $T$ $$\begin{align} T: ((x;v_{1}),(x;v_{2}),(x;v_{3})) \to \frac{\partial }{\partial x} (\omega (x)((x;v_{2}),(x;v_{3}))) \cdot v_{1} \end{align}$$ where $(x;v_{i})_{i=1,2,3}$ are vectors fields.
Now I'm wondering how to prove the operator $T$ is "not" a tensor field on $\mathbb{R}^3$?
Here I first notice that $T$ indeed sents $3$ vectors into a number, since $\omega (x)((x;v_{2}),(x;v_{3}))$ is a number and it only depends on $x$, so the derivative with respect to $x$ is a $1 \times 3$ matrix. Finally, it dot products with $3 \times 1$ column $v_{1}$, the result should be a number.
Then I tried to write out a partial derivative. I think the result of $\omega (x)((x;v_{2}),(x;v_{3}))$ depends on $x$ , $v_{2}$ and $v_{3}$ where the last 2 also depend on $x$ . Thus I think I might be able to write $\omega$ as $$\begin{align} \omega(x,v_{2}(x) ,v_{3}(x)) = r \end{align}$$ Thus the partial derivative is $$\begin{align} \frac{\partial \omega}{\partial x}+ \frac{\partial \omega}{\partial v_{2}} \cdot \frac{\partial v_{2}}{\partial x}+ \frac{\partial \omega }{\partial v_{3}} \cdot \frac{\partial v_{2}}{\partial x} && (*) \end{align}$$ where $\dfrac{\partial \omega}{\partial v_{2}} \cdot \dfrac{\partial v_{2}}{\partial x}$ can be regarded as a $1 \times 3$ matrix $$\begin{align} \left( \frac{\partial \omega}{\partial v_{2}^1} \ \frac{\partial \omega}{\partial v_{2}^2} \ \frac{\partial \omega}{\partial v_{2}^3} \right) \end{align}$$ and $\dfrac{\partial v_{2}}{\partial x}$ is a $3 \times 3$ matrix, and hence $(*)$ is indeed a $1 \times 3$ matrix.
So far, I have 2 questions
1 Are my thoughts above right?
2 If they are right, how can I tell that $T$ is "not" a tensor field? Any help or hints on this? Thanks!
Note: I found this one quite hard to explain, and ended up making two major edits.
You've done a great job of navigating the definitions. You're absolutely right that a (covariant) $2$-tensor is a function that sends two tangent vectors to a number, and you're also right that a tensor field is a function that sends a point to a tensor. It took me a lot of time to get my head around these definitions, so if you're only starting out, then this is really good progress.
Overall, I think the notation could be cleaned up a bit. In particular, the use of the notation $\omega(x, v_1(x), v_2(x))$ to represent the action of the tensor $\omega|_x$ on the vectors $v_1|_x$ and $v_2|_x$ seemed to have thrown you off. That notation suggests that this quantity is a function of three variables, $x$, $v_1$ and $v_2$ - hence why you applied the chain rule when you calculated the derivative on the next line. In really, when we take the derivative with respect to $x$, we should view $v_1$ and $v_2$ as fixed. We shouldn't be differentiating $\omega$ with respect to $v_1$ and $v_2$.
Furthermore, you have clearly grasped the idea that a vector field has a component representation. The next step is then to get familiar with the idea that a tensor field has a component representation too. This idea is key to solving the problem.
Let me show you...
If $v$ is any smooth vector field on $\mathbb R^3$ and $x$ is any point in $\mathbb R^3$, then $v|_x$ is a vector in the tangent space $\mathcal T_x(\mathbb R^3)$. We can write $v|_x$ in terms of its components, giving us an expression like $$v|_x = (v^1(x), v^2(x), v^3(x)).$$ The functions $v^1(x), v^2(x), v^3(x)$ represent the components of $v|_x$, at various different points $x \in \mathbb R^n$. The $v^i(x)$'s are smooth functions of $x$, since $v$ is smooth vector field.
Similarly, if $\omega$ is a smooth $2$-tensor field on $\mathbb R^3$ and $x$ is a point in $\mathbb R^n$, then $\omega|_x$ is a $2$-tensor acting on the tangent space $\mathcal T_x(\mathbb R^3)$. If $v_1$ and $v_2$ are two vector fields on $\mathbb R^3$, then we have an expression like $$\omega|_x(v_1|_x, v_2|_x) = \sum_{i = 1}^3\sum_{j=1}^3\omega_{ij}(x)v_1^i(x)v_2^j(x),$$ where the functions $\omega_{ij}(x)$'s represent the components of $\omega|_x$ at various points $x \in \mathbb R^n$. The $\omega_{ij}(x)$'s are smooth functions of $x$, since $\omega$ is smooth.
If you're wondering why such an expression exists, then remember that $\omega|_x$ acts bilinearly on vectors in $\mathcal T_x(\mathbb R^n)$. The bilinearity of $\omega|_x$ is part of the definition of what it means for $\omega|_x$ to be a tensor. A standard result in linear algebra says that if $V$ is any vector space and $f : V \times V \to \mathbb R$ is a bilinear map, then there exist numbers $f_{ij}$ such that $f(v_1, v_2) = \sum_{i = 1}^3\sum_{j = 1}^3 f_{ij}v_1^i v_2^j$ for any vectors $v_1 = (v_1^1, v_1^2, v_1^3)$ and $v_2 = (v_2^1, v_2^2, v_2^3)$; the numbers $f_{ij}$ are the components of the bilinear map $f$. To obtain the above component expression for $\omega|_x(v_1|_x, v_2|_x)$, just apply this linear algebra fact to $\omega|_x$, for each $x \in \mathbb R^n$.
Anyway, the partial derivative of $\omega|_x(v_1|_x, v_2|_x)$ with respect to a coordinate $x_k$ is given by the expression \begin{multline} \frac{\partial}{\partial x^k}\left( \omega|_x(v_1|_x, v_2|_x) \right) = \\ \sum_{i = 1}^3\sum_{j=1}^3 \left( \frac{\partial \omega_{ij}}{\partial x^k}(x)v_1^i(x)v_2^j(x) + \omega_{ij}(x)\frac{\partial v_1^i}{\partial x^k}(x)v_2^j(x) + \omega_{ij}(x)v_1^i(x)\frac{\partial v_2^j}{\partial x^k}(x) \right).\end{multline} Notice that I'm using the product rule of differentiation, not the chain rule!
Thus, your operator $T$ is described by the formula \begin{multline} T|_x(v_1, v_2, v_3) = \\ \sum_{i = 1}^3\sum_{j=1}^3\sum_{k = 1}^3 \left( \frac{\partial \omega_{ij}}{\partial x^k}(x)v_1^i(x)v_2^j(x)v_3^k(x) + \omega_{ij}(x)\frac{\partial v_1^i}{\partial x^k}(x)v_2^j(x)v_3^k(x) + \omega_{ij}(x)v_1^i(x)\frac{\partial v_2^j}{\partial x^k}(x)v_3^k(x) \right).\end{multline}
Our task is to "show that $T$ is not a $3$-tensor field". Let's figure out how to interpret this task.
I hope you can see that the sentence "Show that $T$ is not a $3$-tensor field" is sloppy use of language - it is a bit like comparing apples to oranges. However, it's clear what the author's intention is. When the author asks you to "show that $T$ is not a $3$-tensor field", what the author really means wants to you to do is:
Okay, so if there did exist such a $3$-tensor field $\lambda$, then for any $x \in \mathbb R^3$, then $$ T|_x(v_1, v_2, v_3) = T|_x(v'_1, v'_2, v'_3)$$ must hold for any vector fields $v_1, v_2, v_3, v'_1, v'_2, v'_3$ such that $$v_1|_x = v'_1|_x, \ \ \ \ \ v_2|_x = v'_2|_x, \ \ \ \ \ v_3|_x = v'_3|_x.$$
But this is not the case. Our expression for $T|_x(v_1, v_2, v_3)$ shows that $T|_x(v_1, v_2, v_3)$ has some dependence on the partial derivatives $\frac{\partial v_1^i}{\partial x^k}(x)$ and $\frac{\partial v_2^j}{\partial x^k}(x)$, in addition to its dependence on $v_1^i(x)$, $v_2^j(x)$ and $v_3^k(x)$. It is possible for vector fields $v_1$ and $v'_1$ to agree at $x$ (i.e. $v_1^i(x) = (v'_1)^i(x)$), and yet have different partial derivatives at $x$ (i.e. $\frac{\partial v^i_1}{\partial x^k}(x) \neq \frac{\partial (v'_1)^i}{\partial x^k}(x)$ for some $k$). Thus it is possible for $T|_x(v_1, v_2, v_3)$ to be different from $T|_x(v'_1, v_2, v_3)$, even when $v_1|_x$ is equal to $v'_1|_x$.
If you found that hard to digest, then here's a different way to think about it. Suppose that there exists a $3$-tensor field $\lambda$ such that $T|_x(v_1, v_2, v_3) = \lambda|_x(v_1|_x, v_2|_x, v_3|_x)$ for all $x \in \mathbb R^3$ and for all vectors fields $v_1, v_2, v_3$. Then it would be possible to write down an expression of the form $$ T|_x(v_1, v_2, v_3) = \sum_{i = 1}^3\sum_{j = 1}^3\sum_{k = 1}^3 \lambda_{ijk}(x) v_1^i (x) v_2^j(x) v_3^k(x).$$ But our $T|_x(v_1, v_2, v_3)$ cannot be expressed in this manner, due to its dependence on the partial derivatives $\frac{\partial v_1^i}{\partial x^k}(x)$ and $\frac{\partial v_2^j}{\partial x^k}(x)$.