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Let $f:M\to \Bbb R^K$ be a smooth function, $\nabla$ be the Euclidean connection on $\Bbb R^K$.

Is it true that $\nabla_{\partial_i f}\partial_j f = \partial_{ij} f\ $?

Edit: Everything after this point could be nonsense. Please feel free to ignore it and just consider the main question.

What I tried is writing $\partial_i f = \frac{\partial f^k}{\partial x^i} \frac{\partial }{\partial y^k}$ thus $$\begin{align} \nabla_{\partial_i f}\partial_j f &= \left( \frac{\partial f^l}{\partial x^i} \cdot \frac{\partial }{\partial y^l} \frac{\partial f^k}{\partial x^j} \right) \frac{\partial }{\partial y^k} \end{align}$$ but I don't know how to calculate this part $$ \frac{\partial }{\partial y^l} \frac{\partial f^k}{\partial x^j}. $$

Perhaps I'm overlooking something silly. Could someone help me confirming/disproving the statement?

BigbearZzz
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    Why did you write $\partial_i f = \frac{\partial f^k}{\partial x^i} \frac{\partial }{\partial y^k}$? Notice the eqquality does not make sense: on the left hand side you have a function and on the right one a vector field. – Mariano Suárez-Álvarez Mar 30 '18 at 16:52
  • It is true, and you don't even need a connection on $\mathbb{R}^K$; you're just taking the directional derivative of (a smooth extension of) a function $\partial_j f$. – user7530 Mar 30 '18 at 16:55
  • @MarianoSuárez-Álvarez Can't we make the identification at least locally around the point where $df\ne 0$? $f$ is a $K$-variables function so its derivative is identified with a vector field. – BigbearZzz Mar 30 '18 at 16:57
  • @user7530 Could you please explain it in a bit more details? I must admit that I am very bad at differential geometry. – BigbearZzz Mar 30 '18 at 16:58
  • @MarianoSuárez-Álvarez Perhaps the way I write/explain that is very sub-optimal, since I am still quite confused myself about how should I write $\partial_i f$ properly. I want to identify $\partial_i f(x)$ with a tangent vector at the point $f(x)\in \Bbb R^K$. – BigbearZzz Mar 30 '18 at 17:04
  • Some of your sums are over indices going from 1 to $\dim M$, and others go from 1 to $K$? – Mariano Suárez-Álvarez Mar 30 '18 at 17:28
  • @MarianoSuárez-Álvarez All of the sums go from $1$ to $K$. $i$ and $j$ are fixed. I used $x$ for coordinates in $M$ and $y$ for $\Bbb R^K$. – BigbearZzz Mar 30 '18 at 17:36
  • I still don't understand what $\partial_i f$ means, if it's supposed to be a vector. – user7530 Mar 30 '18 at 17:50
  • @user7530 I want to think of it as the vector $\partial_i f(x)=[\partial_i f^1(x),\dots, \partial_i f^k(x)] \in \Bbb R^k$, where $x=(x^1,\dots,x^m)\in M$. – BigbearZzz Mar 30 '18 at 18:19
  • More precisely, $\partial_i f(x)\in T_{f(x)}\Bbb R^k$. – BigbearZzz Mar 30 '18 at 18:23
  • I'm not reading all the stuff here, but you need to differentiate in the direction of a tangent vector to $M$, which $\partial_i f$ most certainly is not. – Ted Shifrin Mar 31 '18 at 04:36

1 Answers1

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taking this up with respect to the previous discussion here Seeing that the second fundamental form is the orthogonal component of the Laplacian, I think you may have some confusion.

What you are asking for this question, that $\nabla_{\partial_i f}{\partial_jf}=\partial_i\partial_j f $, is certainly not true in general. Recall that for general coordinates ${x^1\ldots x^m}$ on a manifold $M$, the second derivative does not have any intrinsic meaning! The point of the covariant derivative is that it is intrinsically defined. $\nabla \colon TM \times TM \to TM$. So it always takes two tangent vectors and spits out a new tangent vector. So the left hand side in your expression is perfectly covariant (i.e. it is a tangent vector) whereas the right hand side is not, it may well vanish completely in one coordinate system and not some other (something certainly not true of a tangent vector!)

Now, there is one specific case in which your identity holds, and that is when we deal with a flat ambient connection, which was the context of the previous question.

To illustrate that, suppose that $M$ is a submanifold of Euclidean space $\mathbb{R}^n$, and we will denote the flat connection on Euclidean space by $\bar{\nabla}$. Now let $\{y^1, \ldots, y^n\}$ be Cartesian coordinates on $\mathbb{R}^n$. That $\bar{\nabla}$ is flat means $\bar{\nabla}_{\frac{\partial}{\partial y^\alpha}} \frac{\partial}{\partial y^\beta}=0$. Let $f\colon M \to \mathbb{R^n}$ be the embedding of $M$ and let $\{x^1 ,\ldots , x^m\}$ be some local coordinates on $M$. Now the $\frac{\partial f}{\partial x^i}$ form a local frame for the tangent space to $M$. Then by the rules of the connection $$ \bar{\nabla}_{\frac{\partial f}{\partial x^i}} \frac{\partial f}{\partial x^j} = \bar{\nabla}_{\frac{\partial f^\alpha}{\partial x^i} \frac{\partial}{\partial y^\alpha}} ( \frac{\partial f^\beta}{\partial x^j} \frac{\partial}{\partial y^\beta} ) = \frac{\partial f^\alpha}{\partial x^i} \bar{\nabla}_{\frac{\partial}{\partial y^\alpha}} ( \frac{\partial f^\beta}{\partial x^j} \frac{\partial}{\partial y^\beta} ) \\ =\frac{\partial f^\alpha}{\partial x^i} (\frac{\partial^2 f^\beta}{\partial y^\alpha \partial x^j} \frac{\partial}{\partial y^\beta} + \frac{\partial f^\beta}{\partial x^j}\bar{\nabla}_{\frac{\partial}{\partial y^\alpha}} \frac{\partial}{\partial y^\beta}) = \frac{\partial f^\alpha}{\partial x^i} \frac{\partial^2 f^\beta}{\partial y^\alpha \partial x^j} \frac{\partial}{\partial y^\beta} \\ =\frac{\partial^2 f^\beta}{\partial x^i \partial x^j} \frac{\partial}{\partial y^\beta} = \frac{\partial^2 f}{\partial x^i \partial x^j} $$ where in moving to the last line we applied the chain rule.

Aerinmund Fagelson
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    You write that the covariant derivative takes "two tangent vector..." but it doesn't: one at least had to be a field. – Mariano Suárez-Álvarez Mar 30 '18 at 22:37
  • Good point! Thank-you :) whenever I wrote tangent vector what I really meant was tangent field. But you are right that it only really requires one tangent field (to be differentiated), and the value of a tangent field at a given point, (a tangent vector if you prefer, in which direction to differentiate). – Aerinmund Fagelson Mar 30 '18 at 22:41
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    @AerinmundFagelson Thank you for taking you time, I really appreciate it. Still there's a part that confuses me, i.e. how do we define $\nabla_{\partial_i f}\partial_j f$ rigorously when $f$ is not an embedding, especially at the points where $f$ is not injective or $df$ fails to have full rank. – BigbearZzz Mar 30 '18 at 23:27
  • I think in the context of harmonic map, $u$ is merely a smooth function $u:M\to N$ as well. – BigbearZzz Mar 30 '18 at 23:28
  • Ah! Well I think you are right that $f$ need not be an embedding, and not even an immersion to justify the above (as may be useful for your harmonic map). In any case the $\frac{\partial f}{\partial x^i}$ are tangent vectors in $\mathbb{R}^n$, and you can hit them with the ambient connection. Of course as geometers we would prefer $f$ to be an embedding or an immersion. If $f$ fails to be an embedding then $f(M)$ need not be a submanifold, and if $f$ fails to be an immersion then we can't even talk define extrinsic quantities such as the second fundamental form, mean curvature etc. of $f(M)$. – Aerinmund Fagelson Mar 31 '18 at 09:38
  • My apologies, my last comment was not satisfactory. Thinking over my answer again I realise there is a point that I failed to observe, regarding when we can "hit vectors with the ambient connection" as I blithely put it. Namely, the $\frac{\partial f}{\partial x^i}$ are technically not vector fields in the ambient space $\mathbb{R}^n$, but are only defined along $f(\Sigma)$, so you can't just hit them with the ambient connection $\bar{\nabla}$ (forgetting my own golden rule, checking where everybody lives!). – Aerinmund Fagelson Apr 10 '18 at 16:44
  • Usually, when $f$ is an immersion, we work as follows. For all $p\in \Sigma$ there exists a neighbourhood $U$ of $p$ such that $f(U)$ is a submanifold, and so we can find a tubular neighbourhood $N$ of $f(U)$. Then, letting $X_i=\frac{\partial f}{\partial x^i}$ we can always find extensions $\tilde{X}i$ of the $X_i$ defined on $N$, and we define $(\bar{\nabla}{X_i} X_j) = \bar{\nabla}_{\tilde{X}_i} \tilde{X_j}$ at $p$, and since the $X_i$ are tangent to $f(U)$, its simple to show that this is independent of the choice of extension, so well defined. – Aerinmund Fagelson Apr 10 '18 at 16:49
  • But this argument breaks down when $f$ is not an immersion, and I guess this is the point you were getting at when you ask how to define $\bar{\nabla}_{\partial_i f} {\partial_j f}$ rigorously at those points. Frankly I'm a little bit stuck by that now as well! Perhaps its still possible to construct an extension, but I'm struggling to do it rigorously.. – Aerinmund Fagelson Apr 10 '18 at 16:58
  • The only point I can think of now, in the specific case of harmonic maps, is that the harmonic map equation can still be posed even if $f$ is not an immersion, so perhaps it is enough for the purpose of PDE just to derive that equation for smooth immersions (a nice class of functions), then forget about the geometry and simply define a harmonic map as something solving that PDE, even if the solution is some weaker object which can't be so easily interpreted geometrically. I think that is quite often the moral anyway... – Aerinmund Fagelson Apr 10 '18 at 17:08
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    The general way to do these things rigorously is using the pullback connection $f^* \tilde \nabla,$ which acts naturally on sections of the vector bundle $f^* T \mathbb R^n$, otherwise known as "vector fields along $f$". – Anthony Carapetis Apr 25 '18 at 13:55
  • Do you have a reference for that construction? As I understand it the pull-back connection would be defined something like $(f^\nabla)_X f^s = f^*( \nabla_{df(X)} s)$, but how do you make sense of the right hand side without defining some extension of $df(X)$. It's that part which I'm not sure how to do when $f$ is not an immersion. – Aerinmund Fagelson Apr 26 '18 at 14:53
  • @AerinmundFagelson: remember that $\nabla_XY|_p$ depends on $X$ only via the pointwise value $X_p;$ so there's no extension necessary. – Anthony Carapetis Apr 30 '18 at 01:10
  • @AnthonyCarapetis: Sorry my last comment was not properly thought out. I actually have difficulty as to how you define the vector field to be differentiated. To compute $\nabla_{\partial_i f} {\partial_j f}$, we only need $\partial_i f$ defined along the image of $f$. But we need to define an extension of $\partial_j f$. Since the pullback connection will only act on sections $f^*(s)$ where $s$ is a section defined on $N$, how do we extend $\partial_j f$ in our example to a section $s$? – Aerinmund Fagelson May 02 '18 at 14:48
  • @AerinmundFagelson: the pullback connection acts on any section; it's just defined in terms of pullback sections. You can write a general section in the form $s=\sum_i \phi_i f^* s_i$ where the $\phi_i$ are scalar functions and the $s_i$ are sections on $N$, and then use the linearity/Leibniz properties of connections to expand $(f^* \nabla)(s)$ this so that the connection acts only on the $f^* s_i.$ – Anthony Carapetis May 03 '18 at 00:31
  • To be practical, the pullback connection is easiest to work with in coordinates, where (if we let $\Gamma^a_{bc}(y)$ denote the coefficients/Christoffel symbols of $\tilde \nabla$ on $N$) it's simply the connection with coefficients $\omega^a_{bk}(x) = \sum_c \Gamma^a_{bc}(f(x)) \partial_k f^c(x).$ – Anthony Carapetis May 03 '18 at 00:35