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Motivated by this question, is there a $n$ dimensional Riemannian manfold $M$, $n>1$ such that the space of all gradient vector fields is a Lie algebra under the usual Lie bracket of vector fields.

For $n=1$ there is an example and therev is a no-example: The example is the real line and the counter example is $S^1$.

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    $\begingroup$ You can easily adapt the example for $S^1$ when $M$ has dimension $> 1$. Here is the sketch: take any point $p \in M$ put polar coordinates in $T_pM$ such that $\theta$ is one angular coordinate and compose with the exponential map at $p$ to get a local chart near $p$. Then consider the functions $\sin(\theta)$ and $\cos(\theta)$ and get a contradiction by observing the behaviour of the Lie bracket of their gradients on a circle with parameter $\theta$. $\endgroup$
    – Holonomia
    Apr 1, 2017 at 20:21
  • $\begingroup$ @Holonomia May I ask you to elaborate your comment in the form of an answer.. Thank you. $\endgroup$ Apr 2, 2017 at 2:08
  • $\begingroup$ Which class of functions are you allowing here? If you allow all $C^\infty$ functions, then this can be checked locally and is impossible for $n > 1$. If you require the functions to be algebraic, the answer can be more complex. $\endgroup$
    – user44191
    Apr 2, 2017 at 2:41
  • $\begingroup$ @user44191 I consider smooth functions. But the metric is arbitrary. For arbitrary metric, why is it impossible, locally?)n>1) $\endgroup$ Apr 2, 2017 at 2:45
  • $\begingroup$ I'm not at a computer, so it would be hard to write it out fully, but the idea is that gradients have to satisfy certain differential conditions (corresponding to the fact that the exterior derivative squares to 0) that Lie brackets of gradients don't. That can be used to show that given a point $p_0$, a tangent vector $\vec{v} \in T_{p_0}$, and an element of $T^2_{p_0}$, there are some functions $f, g$ which give a gradient-bracket that matches those at $p_0$, while a gradient must satisfy certain conditions on the last part. $\endgroup$
    – user44191
    Apr 3, 2017 at 3:07

2 Answers 2

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Let $(M^n,g)$ $n>1$ a Riemannian manifold and let $i: S^1 \to M$ a embedding of $S^1$ (it is easy to see that there are lots of such embeddings, of course here you use $n>1$...). Then take a tubular neighborhood of $i(S^1)$ as explained in Theorem 1 here:

http://montgomery.math.ucsc.edu/classes/mfds/normalbundle.pdf

Then use the coordinate $\theta$ on $i(S^1)$ and $x_1,x_2,\cdots,x_{n-1}$ for the normal directions in the tubular neighborhood. Now the functions $\sin(\theta)$ and $\cos(\theta)$ are defined on the tubular neighborhood (they do not depend upon $x_1,\cdots,x_{n-1}$). Now compute the gradient of $\sin(\theta)$ and $\cos(\theta)$ and get the same contradiction as in the question you posted before.

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  • $\begingroup$ I'm not so sure that gets the same contradiction as before; the extra dimensions can add some interesting interactions to the Lie bracket. Why would the Lie bracket necessarily have a nonzero loop integral? $\endgroup$
    – user44191
    Apr 5, 2017 at 17:17
  • $\begingroup$ Just compute the gradients and see that the bracket of the gradients restricted to $i(S^1)$ is the vector field $\frac{\partial}{\partial \theta}$. Now you should see the nonzero loop integral . Should I edit my answer and add this last few lines? $\endgroup$
    – Holonomia
    Apr 5, 2017 at 17:23
  • $\begingroup$ No; I think I see what I was missing. I missed the Riemannian part of the link for tubular neighborhood, and missed that the $x_i$ would be orthogonal to the circle. $\endgroup$
    – user44191
    Apr 5, 2017 at 17:54
  • $\begingroup$ @Holonomia Thank you very much for your very interesting answer.What can be said about the Lie algebra generated by Gradient vec. fields? Please see my comment to the other answer. $\endgroup$ Apr 5, 2017 at 19:09
  • $\begingroup$ @Holonomia It is difficult to choose one answer to accept. Each answer has its own interesting point. Thank you again for your answer. meta.mathoverflow.net/questions/1491/… $\endgroup$ Apr 5, 2017 at 19:14
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Expanding on my comment: I will show that even looking only locally, this is impossible for dimension $n > 1$.

Define $\pi: TM \rightarrow T^*M$ as the "lowering the index" operator, taking a vector field $X \in TM$ to the covector field $\alpha \in T^*M$, where $\alpha(Y) = \langle X, Y\rangle$ for any vector field $Y$. This operator $\pi$ induces a natural linear operator from $\Gamma(TM)$ to $\Gamma (T^{*} (M)$. We again use $\pi$ to denote this induced operator. For the sake of notation, define $\{f, g\} = \langle \nabla f, \nabla g \rangle = (\nabla f)(g) = (\nabla g)(f)$

By definition, $\nabla f = \pi^{-1}(df)$. Therefore, for any gradient, we will have that $d(\pi(\nabla f)) = d(df) = 0$. We therefore only need to show that there are $f, g$ with $d(\pi([\nabla f, \nabla g])) \neq 0$. Note that this is in fact an if-and-only-if if we work locally.

Because the metric is nondegenerate, we can check what happens for vector fields that are gradients. We'll contract with one gradient at a time.

$\iota_{\nabla h}(d(\pi([\nabla f, \nabla g]))) = \mathcal{L}_{\nabla h}(\pi([\nabla f, \nabla g])) - d(\pi([\nabla f, \nabla g])(\nabla h))$

The second term is equal to $d(\langle [\nabla f, \nabla g], \nabla h \rangle)$. By the definition of $\nabla h$, this is equal to $d(\{f, \{g, h\}\} - \{g, \{f, h\}\})$.

$\iota_{\nabla j} \iota_{\nabla h}(d(\pi([\nabla f,\nabla g]))) = \iota_{\nabla j}(\mathcal{L}_{\nabla h}(\pi([\nabla f, \nabla g])) - d(\{f, \{g, h\}\} - \{g, \{f, h\}\}))$

$= \mathcal{L}_{\nabla h}(\pi([\nabla f, \nabla g])(\nabla j)) - \iota_{\mathcal{L}_{\nabla h}(\nabla j)}(\pi([\nabla f, \nabla g])) - \{j, \{f, \{g, h\}\}\} + \{j, \{g, \{f, h\}\}\}$

$= - \langle [\nabla f, \nabla g], [\nabla h, \nabla j]\rangle + \{h, \{f, \{g, j\}\}\} - \{h, \{g, \{f, j\}\}\} - \{j, \{f, \{g, h\}\}\} + \{j, \{g, \{f, h\}\}\}$

So we want to check if this is true for every $f, g, h, j$:

$\langle [\nabla f, \nabla g], [\nabla h, \nabla j]\rangle = \{h, \{f, \{g, j\}\}\} - \{h, \{g, \{f, j\}\}\} - \{j, \{f, \{g, h\}\}\} + \{j, \{g, \{f, h\}\}\}$

Call the left side $A(f, g, h, j)$ and the right side $B(f, g, h, j)$. Assume that $A(f, g, h, j) = B(f, g, h, j)$ for all $f, g, h, j$. Then $A(f, g g', h, j) - g A(f, g', h, j) - g' A(f, g, h, j) = B(f, g g', h, j) - g B(f, g', h, j) - g' B(f, g, h, j)$.

A quick calculation shows that $A(f, g g', h, j) - g A(f, g', h, j) - g' A(f, g, h, j) = \{f, g\} \{h, \{j, g'\}\} - \{f, g\} \{j, \{h, g'\}\} + \{f, g'\} \{h, \{j, g\}\} - \{f, g'\} \{j, \{h, g\}\}$.

On the other hand, $B(f, g g', h, j) - g B(f, g', h, j) - g' B(f, g, h, j) = \{f, g\} \{h, \{g', j\}\} + \{h, g\} \{f, \{g', j\}\} + \{g, j\} \{h, \{f, g'\}\} + \{f, g'\} \{h, \{g, j\}\} + \{h, g'\} \{f, \{g, j\}\} + \{g', j\} \{h, \{f, g\}\} - \{h, g\} \{g', \{f, j\}\} - \{h, g'\} \{g, \{f, j\}\} - (\{f, g\} \{j, \{g', h\}\} + \{j, g\} \{f, \{g', h\}\} + \{g, h\} \{j, \{f, g'\}\} + \{f, g'\} \{j, \{g, h\}\} + \{j, g'\} \{f, \{g, h\}\} + \{g', h\} \{j, \{f, g\}\} - \{j, g\} \{g', \{f, h\}\} - \{j, g'\} \{g, \{f, h\}\})$

So we want to check whether:

$\{h, g\}(- \{g', \{f, j\}\} - \{j, \{f, g'\}\} + \{f, \{g', j\}\}) + \{j, g\}(\{g', \{f, h\}\} + \{h, \{f, g'\}\} - \{f, \{g', h\}\}) + \{h, g'\}(- \{g, \{f, j\}\} - \{j, \{f, g\}\} + \{f, \{g, j\}\}) + \{j, g'\}(\{g, \{f, h\}\} + \{h, \{f, g\}\} - \{f, \{g, h\}\}) = 0$.

Using a similar trick by splitting $f$ to $f f'$, we get:

$2 \{h, g\}\{g', f\}\{f', j\} + \text{similar terms} = 0$, where "similar terms" refers to all terms gotten by switching g and g', switching f and f', and switching h and j and negating.

We've finally made it to a pointwise condition - which means that we only need to check this on a vector space with the standard metric. And it is false; take $h = g = g' = f = x, f' = j = y$. More generally, if we let $h, g, g', f, f', j = a_i x + b_i y$, the equation is:

$(a_1 a_2 + b_1 b_2)(a_3 a_4 + b_3 b_4)(a_5 a_6 + b_5 b_6) + \text{similar terms} = 0$.

The terms consisting of just $a$ or just $b$ cancel out; the only terms left either have 4 $a$s and 2 $b$s or vice versa. It's not hard to see that these terms can't cancel - the two $b$s (or the two $a$s) must remain together, which gives a unique product it can come from.

To summarize: We took the question, extracted a differential condition from it, extracted a pointwise condition from the differential condition, and showed that the pointwise condition could not be satisfied in dimension > 1.

ETA: To give a more explicit answer: choose coordinates $x, y$. Then:

$d(\pi([\nabla x^2, \nabla xy])) = d(\pi([2x \nabla x, x \nabla y + y \nabla x])) = d(\pi(2x \nabla y - 2y \nabla x + 2x^2 [\nabla x, \nabla y]))$

$= d(2x dy - 2y dx + 2x^2 [\nabla x, \nabla y]) = 4 dx \wedge dy + 4x dx \wedge \pi([\nabla x, \nabla y]) + 2x^2 d(\pi([\nabla x, \nabla y]))$

Just by checking at 0, we can see that this is nonzero - so $d(\pi([\nabla x^2, \nabla xy]))$ must be nonzero - so as long as $dx$ and $dy$ are linearly independent at some point where $x = 0$ (hence the need for $n > 1$), then at that point, $[\nabla x^2, \nabla xy]$ is not a gradient.

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  • $\begingroup$ I really thank you very much for your very interesting answer. As we see, the coefficient $g_{ij}$ of metric does not play a crucial role in your interesting answer. So this situation is a vague motivation for the question "Does the Lie algebra structure of the Lie subalgebra generated by gradient vector fields depend on a chosen Riemannian metric"? What is a description of the Lie algebra generated by Gradient vector fields corresponding to a given metric? Thanks again for your help. $\endgroup$ Apr 5, 2017 at 19:07
  • $\begingroup$ It is difficult to choose one answer to accept. Each answer has its own interesting point. Thank you again for your answer. meta.mathoverflow.net/questions/1491/… $\endgroup$ Apr 5, 2017 at 19:15

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