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The two unit vectors

$$\left\{\frac{(-y,x,0)}{\sqrt{x^2+y^2}}, \frac{(-xz,-yz,x^2+y^2)}{\sqrt{x^2+y^2}}\right\}$$

form an orthonormal basis to the tangent space at $(x,y,z)\in \mathbb{S^2}\backslash {(0,0,\pm1)}$. Observe that $$(-xz,-yz,x^2+y^2)=(x,y,z)\times (-y,x,0).$$

Now let $u$ be a smooth function on $\mathbb{S}^2$ and define

$$\nabla u (x,y,z) \cdot \frac{(-xz,-yz,x^2+y^2)}{\sqrt{x^2+y^2}}=\nabla u (x,y,z) \cdot \frac{(-xz,-yz,1-z^2)}{\sqrt{x^2+y^2}}=\nabla u (x,y,z) \cdot \frac{(0,0,1)}{\sqrt{x^2+y^2}},$$ where I have used $\nabla u(x,y,z)\cdot (x,y,z)=0$.

Since $\nabla u (0,0,1)$ is parallel to $xy-$plane, intuitively one expect the projection of $\nabla u$ onto $\frac{(-xz,-yz,x^2+y^2)}{\sqrt{x^2+y^2}}$ to be small as $(x,y,z)\rightarrow 0$, and hence $$\lim_{(x,y,z)\rightarrow (0,0,1)}\nabla u (x,y,z) \cdot \frac{(0,0,1)}{\sqrt{x^2+y^2}}=0.$$

Is the above limit zero? I am unable to prove or disprove it. I would appreciate any hint.

Veselin Dimov
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MathLearner
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  • If $u$ is a function on $S^2$, for starters, what do you mean by $\nabla u(x,y,z)$? Why not work entirely with $du$ as a $1$-form on $S^2$? – Ted Shifrin Aug 26 '24 at 15:49
  • @T Yes, $u$ is a function on $\mathbb{S}^2$. $\nabla u$ is the direction where $u$ has the largest directional derivative. One can define $\nabla u=\nabla u=u^\theta \partial_{\theta}+u^\varphi \partial_{\varphi}$, everywhere except the two points $(0,0,\pm1)$. – MathLearner Aug 26 '24 at 21:14
  • @TedShifrin For definition of the gradient, please see this post: https://math.stackexchange.com/questions/2228747/definition-of-gradient-of-a-function-f-in-riemannian-manifold/2250437#2250437 – MathLearner Aug 26 '24 at 21:15
  • I know what the definition of the gradient is. It is the vector field that corresponds to the $1$-form $du$ using the Riemannian metric to give an isomorphism $T^*M\cong TM$. Your question is much more naturally phrased using the differential rather than the gradient. $\nabla u(p)\cdot v = du(p)(v)$ for every $v\in T_pM$. – Ted Shifrin Aug 26 '24 at 21:24
  • @TedShifrin I see. I am not a differential geometer, and that's why my question is framed in such an elementary way. – MathLearner Aug 26 '24 at 21:46
  • Your definition in terms of $\partial_\theta$ and $\partial_\varphi$ is, however, incorrect. – Ted Shifrin Aug 26 '24 at 22:48

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The answer to your question is no.

Denote by $v_1=\dfrac{(-y,x,0)}{\sqrt{x^2+y^2}}$, $v_2=\dfrac{(-xz,-yz,x^2+y^2)}{\sqrt{x^2+y^2}}$ the orthonormal basis for $T_{(x,y,z)}S^2$, as you've indicated. Write $\nabla u(x,y,z) = Av_1+Bv_2$, where $A$ and $B$ are functions of $(x,y,z)$. Sadly, since the orthonormal basis does not converge as $(x,y,z)\to (0,0,1)$, there's no reason to expect $A$ and $B$ to be well-behaved in the limit. However, since the projection of $(0,0,1)$ onto $T_{(x,y,z)}S^2$ is the vector $\sqrt{x^2+y^2}v_2$, we'll have $\nabla u\cdot (0,0,1) = B\sqrt{x^2+y^2}$, and so the quantity in your limit is simply $B$, and it need not converge at all, let alone to $0$.

Let's take a specific example. Let $u(x,y,z)=x$. Then $\nabla u(x,y,z)$ is the projection of $(1,0,0)$ onto $T_{(x,y,z)}S^2$, and this clearly approaches $(1,0,0)$ as $(x,y,z)\to (0,0,1)$. Let's calculate your limit without your problematic basis vectors: The projection of $(1,0,0)$ onto the tangent plane is $$(1,0,0)-\big((1,0,0)\cdot (x,y,z)\big)(x,y,z) = (1-x^2,-xy,-xz),$$ and $$(1-x^2,-xy,-xz)\cdot \frac{(0,0,1)}{\sqrt{x^2+y^2}} = -\frac{xz}{\sqrt{x^2+y^2}}.$$ This, likewise, has no limit as $(x,y,z)\to (0,0,1)$.

Ted Shifrin
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