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Let $W$ be a $(0,2)$ tensor in the sense that $W: V^{*r} \times V^r \to \mathbb{R}$ on a manifold $M$ with chart $(U,\phi) = (U, (x^1, x^2))$ and define the "coefficients"

$$W^{11} = x^1 + 5\; \text{and}\; W^{ij} = 0 \;\forall(i,j) \neq (1,1)$$

So basically we just need to look at the $(1,1) $ indices and nothing else.

In the chart $(U,\phi)$, I was asked to find $W((x^2)^2dx^1, 3dx^1 + e^{x^1x^2}dx^2)$

Now apparently the answer just treats the terms $(x^2)^2, 3, e^{x^1x^2}$ as scalars and the answer turned out to be

$$W((x^2)^2dx^1, 3dx^1 + e^{x^1x^2}dx^2) = 3(x^2)^2(x^1 + 5)$$

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Now here was what I thought ($[..]$ will be used to evaluation)

\begin{align} W((x^2)^2dx^1, 3dx^1 + e^{x^1x^2}dx^2) &= W^{11}( \partial / \partial x_1 \otimes \partial / \partial x_1 ) [(x^2)^1 dx^1, 3dx^1 + e^{x^1x^2}dx^2)]\\ &=W^11 (\partial / \partial x_1)[(x^2)^2dx^1]( \partial / \partial x_1 )[3dx^1 + e^{x^1x^2}dx^2] \end{align}

Now what I assume the solution did was "pull out" the "constants" $(x^2)^2, 3, e^{x^1x^2}$. But I cannot figure out why we don't apply a "Leibniz" rule like on the product. For example, why don't we do $$(\partial / \partial x_1)[(x^2)^2 dx^1] = (x^2)^21 + 0dx^1$$? Yet we don't do $$(\partial / \partial x_1)[ e^{x^1x^2}dx^2] = e^{x^1x^2}0 + x^2 e^{x^1x^2}dx^2,$$ but instead we should do $$ e^{x^1x^2}\partial / \partial x_1(dx^2) = 0$$

Lemon
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1 Answers1

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When you write $\partial_1 \otimes \partial_1$, you're looking at $\partial_1$ as a tensor field acting on one-forms, which is linear over functions, and taking the tensor product of two such tensors. You're not looking at $\partial_1$ as a differential operator, which acts on functions. Hence $$\partial_1((x^2)^2\,{\rm d}x^1) = (x^2)^2 \require{cancel}\cancelto{1}{\partial_1({\rm d}x^1)}= (x^2)^2.$$

Ivo Terek
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  • So you are telling me $\partial_1 (a dx^1 + bdx^2 + \dots) = a\partial_1(dx^1) = 1$ because $\partial_1$ is "a tensor field acting on one-forms and linear over functions". This is probably a dumb remark, but doesn't the tangent space basis element $ \partial_1$ also act on objects like $e^{x^1x^2}$? – Lemon Aug 21 '17 at 03:09
  • The final result would be $a$, and not $1$. There is an abuse of notation going on here. Let's think of the "static" situation first: if $V$ is a vector space, $v\in V$ and $\phi\in V^$, then $\phi$ acts in $v$ and produces a number $\phi(v)$. When we say that $v$ acts as a tensor over $\phi$ producing the number $v(\phi)\equiv \phi(v)$, we're using the canonical identification $V\cong V^{}$ and using the functional $\hat{v}\in V^{*}$, defined by $\hat{v}(\phi)\doteq \phi(v) $. – Ivo Terek Aug 21 '17 at 03:15
  • Sorry that was my typo, it just went past the 5 min edit. – Lemon Aug 21 '17 at 03:15
  • The only real thing I really don't get is why we treat the "$0-$form like" objects $e^{x^1dx^2}$ as "constants". Because I know we evaluate $\partial_i(dx^j) = \delta_{ij}$ as the double dual basis – Lemon Aug 21 '17 at 03:16
  • Bearing this is mind, $\partial_1$ acts over functions, but we're dealing with his evil twin $\hat{\partial_1}$ (a resident of $T^{**}M$) – Ivo Terek Aug 21 '17 at 03:17
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    Oh, so there is some isomorphism that identifies $\partial_1 \to \hat{\partial}_1$ – Lemon Aug 21 '17 at 03:18
  • Well, a $0$-form takes each point to a $0$-linear alternating functional in the tangent space, that is, a constant. One constant for each point is a function. – Ivo Terek Aug 21 '17 at 03:20
  • Yes!!! ${}{}{}$ – Ivo Terek Aug 21 '17 at 03:20
  • So the $(0,2)$ tensor $W$ is actually a map $T^{}(M) \times T^{}(M) \to \mathbb{R}...$? – Lemon Aug 21 '17 at 03:24
  • $W$ is a $(0,2)$-tensor field: it smoothly associates to each point in the manifold a $(0,2)$-tensor defined in the tangent space to said point. – Ivo Terek Aug 21 '17 at 03:26
  • Argh slightly gone off topic now... – Lemon Aug 21 '17 at 03:32
  • The evil twin $\hat{\partial_1}$ acts on elements in $(T^M)^$. Objects in this double dual are $adx^i$. What arbitrary object(s) do the $a$ here represent? In this case we have $a:= (x_1,x_2) \to f(x_1,x_2)$ (say $f(x_1,x_2) = e^{x^1x^2} $ in this case) – Lemon Aug 21 '17 at 03:35
  • I'm probably lose you now... but you can think of $W\in\Gamma(T^{(0,2)}(TM)) $ as a section of the $(0,2)$-tensor bundle of $M$, that is, $W:M\to T^{(0,2)}(TM)$ such that for all $p\in M$, $W_p\in T^{(0,2)}(T_pM)$. Meaning that each $W_p: (T_pM)^\times (T_pM)^\to \Bbb R$ is a $(0,2)$-tensor. – Ivo Terek Aug 21 '17 at 03:36
  • Lets talk bout something even more elementary here. Let $V$ be a vector space over field $\mathbb{R}$ with basis $(v_i)$ and its dual $V^$ with basis $(w_j^)$ defined $w^j(av_i) = a\delta^j_{i}$ where $a \in \mathbb{R}$. On its double dual, $V^{*}$, it has say basis elements $(z_k)$ defined "canonically" by $z_k(w^j) = \delta_{ij}$. Now in this setting, say I attach "scalar" components to $w^j$, by $bw^j$, obviously we can define $z_k(bw_j) = bz_k(w_j)$. Now my question, what space/"field" do the objects $b $ can belong to? Because in this case, it seems like $b $ can be functions – Lemon Aug 21 '17 at 03:49
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    Do you know what a module over a ring is? It's like a vector space, but with a ring instead of a field (algebraic sense). The collection of tensor fields is a module over the ring of smooth functions. The suitable morphisms are linear over the scalars in the ring (in this case, functions) – Ivo Terek Aug 21 '17 at 09:54
  • Yeah I know what modules are and I think I can vaguely relate this to just regular $1-$forms which are vector fields (tensor fields) like $x\partial_1 + y\partial_2$. I just haven't found any source that defines the basis elements $\partial_1$ with the $\hat{\partial}_1$ double dual identification you mentioned. – Lemon Aug 21 '17 at 10:00
  • The honest answer is: in differential geometry no one cares about identifications that boil down to linear algebra, most of the time. So I doubt that this will be explained neatly in some book. – Ivo Terek Aug 21 '17 at 10:10
  • I am actually somewhat confused now after reading this https://math.stackexchange.com/questions/253829/definition-of-a-tensor-for-a-manifold. It appears we really do identify $\partial_1$ in its tangent space, and not its double dual? – Lemon Aug 22 '17 at 17:01