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I am trying to understand the relationship between a jet and the singularities of a curve. Concretely I am trying to understand the following example:

Let $\gamma(t) = (\cos t , b \sin t)$ where $b > 0$ be an ellipse. Let $f$ be the distance squared function from $({1\over 2},0)$ to $\gamma (t)$ i.e.

$$ f(t) = ({1\over 2} - \cos t)^2 + b^2 \sin^2 t$$

For each $t_0$ consider the $2$-jet at $t_0$ of $f$:

$$ (\sin t_0 + (b^2 -1)\sin 2 t_0)t + (\cos t_0 + 2(b^2 -1)\cos 2 t_0) t^2$$

Writing this as $c_1 t + c_2 t^2$ the points $(c_1, c_2)$ will trace out a curve $\delta$ as $t_0$ varies. This curve will cross the $c_2-$axis at points where $f$ has a singularity $A_{\ge 1}$ at $t_0$ and will pass through the origin where $f$ has an $A_{\ge2}$ singularity at $t_0$. The geometrical interpretation of this is that $f$ has an $A_{\ge1}$ singularity at $t_0$ iff the normal to the ellipse at $\gamma (t_0)$ passes through $({1\over 2},0)$ and an $A_{\ge 2}$ singularity iff $({1\over 2},0)$ is the centre of curvature at $\gamma (t_0)$.

I have absolutely no intuition about why the coefficients of the jet should have the property that they trace out a curve and this curves contains information about the singularities of the original curve. Please could someone help me understand why this is true?

This example can be found here on page 57.

student
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1 Answers1

1

Let $\gamma$ be a regular plane curve, $p_{0}$ a point of the plane, and $$ f(t) = \|\gamma(t) - p_{0}\|^{2} = \bigl(\gamma(t) - p_{0}\bigr) \cdot \bigl(\gamma(t) - p_{0}\bigr). $$ Note that $$ f'(t) = 2\bigl(\gamma(t) - p_{0}\bigr) \cdot \gamma'(t), \tag{1} $$ and $$ f''(t) = 2\bigl[\bigl(\gamma(t) - p_{0}\bigr) \cdot \gamma''(t) + \gamma'(t) \cdot \gamma'(t)\bigr]. \tag{2} $$

Let $T$ be the unit tangent field and $N$ the normal field, and write $\gamma' = vT$, $T' = \kappa N$.

By (1), $f$ has a critical point at $t_{0}$ if and only if $\gamma'(t_{0})$ is orthogonal to the displacement $\gamma(t_{0}) - p_{0}$, if and only if $\gamma(t_{0}) - p_{0}$ is a multiple of $N$; by (2), $f''(t_{0}) = 0$ in addition if and only if $$ \bigl(\gamma(t_{0}) - p_{0}\bigr) \cdot \gamma''(t_{0}) + \gamma'(t_{0}) \cdot \gamma'(t_{0}) = 0. $$

Since $\gamma''(t_{0}) \cdot N(t_{0}) = \kappa(t_{0}) v(t_{0})^{2}$, the preceding equation gives $$ \left\lVert \gamma(t_{0}) - p_{0} \right\rVert \kappa(t_{0}) v(t_{0})^{2} = \bigl(\gamma(t_{0}) - p_{0}\bigr) \cdot \gamma''(t_{0}) = -v(t_{0})^{2}, $$ or $\left\lVert \gamma(t_{0}) - p_{0} \right\rVert = -1/\kappa(t_{0})$. That is, $$ \gamma(t_{0}) - p_{0} = \left\lVert \gamma(t_{0}) - p_{0} \right\rVert N(t_{0}) = -\frac{1}{\kappa(t_{0})} N(t_{0}),\quad\text{or}\quad p_{0} = \gamma(t_{0}) + \frac{1}{\kappa(t_{0})} N(t_{0}), $$ which means $p_{0}$ is the center of curvature.

The connection with $2$-jets is immediate: the $2$-jet of $f$ at $t_{0}$ is $$ f(t_{0}) + f'(t_{0})(t - t_{0}) + \tfrac{1}{2} f''(t_{0})(t - t_{0})^{2} = c_{0} + c_{1}(t - t_{0}) + c_{2} (t - t_{0})^{2}. $$ The curve $\bigl(c_{1}(t_{0}), c_{2}(t_{0})\bigr) = \bigl(f'(t_{0}), \frac{1}{2}f''(t_{0})\bigr)$ crosses the $c_{2}$-axis if and only if $c_{1}(t_{0}) = 0$, if and only if $f$ has a critical point at $t_{0}$; the curve touches the origin if and only if both components are zero, if and only if $p_{0}$ is the center of curvature of $\gamma$ at $\gamma(t_{0})$.

Andrew D. Hwang
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