# Edit:

It seems that the link "https://cms.math.ca/Events/Toulouse2004/abs/ss7.html#lt" which contains a talk by Loic Teyssier about homological equations and vanishing cycles is temporally inactive.

I asked this question at MSE but I did not get any answer so I ask it here at MO:

Assume that $M$ is a smooth $n$ dimensional manifold with $n>1$.

Is there a lie algebra structure on $\chi^{\infty} (M)$, the space of all smooth vector fields on $M$, such that we have the following property:

For every vector field $X\in \chi^{\infty}(M)$ the operator $ad_X:\chi^{\infty}(M) \to \chi^{\infty}(M)$ with $ad_{X}(Y)=[X,Y]$ is an elliptic differential operator of positive order (non zero order) when we restrict it to non singular points of $X$.

A motivation for this question is the consideration of the adjoint operator by Loic Teyssier in the following talk. But I do not know if this talk imply that there is a relation between the number of limit cycles and the adjoint operator?

Is there a pre print or published paper extracting from this talk?Is there a published conference proceeding for the corresponding program in Toulouse? What is the role of the adjoint operator in investigation of the number of limit cycles?

This talk is indicated here but it seems that it is inactive now.

https://cms.math.ca/Events/Toulouse2004/abs/ss7.html#lt

My initial motivation for consideration of diff. operators associated with a vector field is the following note:

https://arxiv.org/abs/math/0408037

The note has a false part: " It is claimed that the codimension of the range of derivation operator is equal to the number of limit cycles" but the true version is that "This codimension is an upper bound for the number of limit cycles". The true part of the "Proof" of this note is that :"around a hyperbolic limit cycle, one can solve the PDE $X.f=g$ provided the integral of $g$ along the limit cycle be equal to zero. Another true part of the note is included in Remark 1, which actually contains a proof of the fact that the codimension of $D_X$ is an upper bound for the number of limit cycles" But after 13 years, I think that my dream of finite codimensionality is going to be collapsed. The reason is the following interesting comment by Lukas Geyer

https://math.stackexchange.com/questions/1163800/elliptic-and-fredholm-partial-differential-operators

I understand that my old dream is collapsed, because perhaps his argument, in the above MSE link, for two singularities can be repeated for coexistence of one singularity and one limit cycle. Assume that a limit cycle $\gamma$ surrounds a singularity. Consider the space of all smooth (or analytic) functions vanishing at $\gamma$ and singularity. Then perhaps in this function space we can separate all orbits in the interior of $\gamma$ by $\int_{-\infty}^{+\infty} g(\phi_{t})(x)dt$ hence we have infinite codimension. So I search for some other diff. operators associated with a vector field whose some operator theoretical quantities have some dinamical interpretation. For example fredholm index as a quantity which interprate the number of attractors. This is a motivation to ask for some elliptic operator in the form of $ad_X$ for some other Lie structures on $\chi^{\infty}(\mathbb{R}^2)$.

# Note:

The codimension of the range of differential operators for algebraic vector field is introduced here

Codimension of the range of certain linear operators

• So you want this Lie Bracket to be completely unrelated to the existing Lie bracket on vector fields? Also, should it be $\mathbb{R}$-linear with respect to the existing vector space structure on $\chi^\infty(M)$? – Michael Bächtold Jun 5 '17 at 9:16
• @MichaelBächtold As your first question, we do not require any relation with the usual Lie Bracket structure.But we hope that the $ad_X$ operator, would be an elliptic operator.Regarding the second question, I think it is included in the definition of Lie algebra that Lie bracket is $\mathbb{R}$-linear. right? – Ali Taghavi Jun 5 '17 at 10:51
• and we hope that the index of this elliptic operator $ad_X$ has some dynamical interpretation. – Ali Taghavi Jun 5 '17 at 10:57
• Have you tried looking at examples where the Lie bracket is a 2nd order differential operator? – Deane Yang Jul 9 '17 at 17:14
• What definition of ellipticity do you want? That the principal symbol doesn't vanish on the slit (co)tangent bundle? – Joonas Ilmavirta Oct 27 '17 at 11:43

Consider the special case $M=\mathbb R^2$ and $X\equiv(1,0)$. Now, given any vector field $Y$ in terms of component functions as $Y(x_1,x_2)=(y_1(x_1,x_2),y_2(x_1,x_2))$, a simple calculation gives $$ad_XY = (\partial_1y_1,\partial_1y_2).$$ This means that $ad_X=\partial_1\otimes I_2$. Here $I_2:\mathbb R^2\to\mathbb R^2$ is the identity operator. (In the Euclidean space we can differentiate somewhat carelessly and freely identify tangent spaces with their duals.)

Given any $x\in\mathbb R^2$ and $\xi\in T_xM$, the principal (and full) symbol is $$\sigma_{ad_X}(x,\xi) = i\xi_1\otimes I_2 : \mathbb R^2\to\mathbb R^2.$$ This is invertible if and only if $\xi_1\neq0$. In particular, at any point $x$ there are non-zero tangent vectors $\xi$ (e.g. $(0,1)$) for which the principal symbol is not invertible (it vanishes entirely!). Therefore the operator is not elliptic in the sense of invertible principal symbols.

This special case is actually not special. The same argument works in any $\mathbb R^n$ or an open subset. If $X$ is a non-vanishing vector field in an open subset of a manifold, you can set up local coordinates so that $X$ becomes a constant vector and the same argument goes through.

• My apology. I forget to give the bounty on time – Ali Taghavi Oct 31 '17 at 10:27
• @AliTaghavi No problem. If you want, you can always start a new bounty, but it's all up to you. – Joonas Ilmavirta Oct 31 '17 at 16:29
• My apology for unaccepted after my previous acceptance. Now I read again your answer. I realize that you are talking about the usual Lie structure not an arbitrary structure. Am I mistaken? – Ali Taghavi Jan 21 '18 at 20:50
• But I appreciate your attention to my question. – Ali Taghavi Jan 21 '18 at 20:51
• @AliTaghavi Yes, I am talking about the usual Lie structure, not an arbitrary one. Upon rereading your question, I think I should have been more general. But I hope this was of some use. – Joonas Ilmavirta Jan 22 '18 at 5:36

Any Lie bracket structure that is a first order differential operator is, by bilinearity and skew-symmetry, of the form $$\mathrm{ad}_X Y = A^{kp}_{ij}(X^i\partial_pY^j- Y^i\partial_pX^j)e_k,$$ where $e_1, \dots, e_n$ is the standard basis of $\mathbb{R}^n$ and we can assume, by skew-symmetry, that $A^{kp}_{ij} = -A^{kp}_{ji}$. For example, the standard Lie bracket is when $$A^{kp}_{ij} = \frac{1}{2}(\delta^i_p\delta^j_k - \delta^j_p\delta^i_k)$$ The symbol of the general adjoint operator is $$\sigma_X(\xi)Y = A^{kp}_{ij}X^i\xi_pY^je_k = \frac{1}{2}A^{kp}_{ij}\xi_p(X^iY^j-X^jY^i)e_k$$ The vector $Y = X$ always lies in the kernel of this symbol.

• That's fine. The other answer was first. – Deane Yang Oct 31 '17 at 16:53
• I am sorry that I unaccepted your answer. I think that you are imposing some restriction on a general Lie algebra structure on $\chi^{\infty} (\mathbb{R}^n)$. Yes? – Ali Taghavi Jan 22 '18 at 6:30
• What restriction? – Deane Yang Jan 28 '18 at 19:43
• you assume that the operator is "first order". Is it always true, for arbitrary lie algebra structure? – Ali Taghavi Jan 28 '18 at 20:31
• Yes, I do assume that it's first order. Do you know how to define a Lie bracket that is higher order? – Deane Yang Jan 29 '18 at 5:02