Lecture
Our immediate goal is to bring germs of curves of the form (6.1) at critical points to the simplest possible form, the so-called «normal form», by changing the parameter $ and choosing
suitable local coordinates on the plane (x, y). Without loss of generality we shall assume that the critical point corresponds to the value t = φ(0) = ψ(0) = 0. Furthermore, suppose
that both functions φ, ψ have finite multiplicity at zero. Then,
x = φ(t), y = ψ(t), φ(0) = ψ(0) = 0, (6.5)
with some integers m, n ≥ 2 and smooth functions φ(t) and ψ(t).
We shall further restrict ourselves to the case m = n + 1, i.e. we shall study
germs of curves
x = φ(t), y = ψ(t), φ(0), ψ(0) ≠ 0, (6.6)
at zero. Formula (6.6) admits the following useful simplification:
PROBLEM 6.2. Show that, by means of a smooth change of the parameter t and a linear change of the variables x, y in formula (6.6), one can
achieve φ(t) = t and ψ(t) = 1 + g(t), or ψ(t) = 1 + g(t) and
φ(t) = t, where g(t) is a smooth function, g(0) = 0.
Using this simplification, one can solve the following, more difficult problems.
PROBLEM 6.3. Prove that, by means of a smooth change of the parameter t and a smooth change of the variables (x, y), the germ of curve (6.6) with exponent n = 2 is brought to the normal form (6.3) with n = 2.
Solution. Without loss of generality we may assume that φ(0) > 0
(if this is not so, make the substitution x ↦ —x). Then the change of parameter
t ↦ t/φ(t) brings the curve (6.6) to the form
x = t, y = tψ(t), ψ(0) ≠ 0.
Using Lemma 5.3, we have ψ(t) = f(t²) + tg(t²). Then the germ of the curve takes the form
x = t, y = t²f(t²) + t·tg(t²) = t²f(t²) + t²g(t²). (6.7)
The change of variable
y ↦ y — t²g(t²)
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brings the germ of the curve (6.7) to the form (6.3) with exponent n = 2. Note that g(0) = f(0) = 0, so all the changes of
variables we have used are smooth local diffeomorphisms on the corresponding spaces. |
PROBLEM 6.4. Prove that, by means of a smooth change of the parameter t and a smooth change of the variables (x, y), the germ of curve (6.6) with exponent n = 3 is brought to the normal form (6.3) with n = 3.
Solution. By Problem 6.2, consider the germ of curve x = t,
y = t(1 + g(t)), where g(0) = 0. By Problem 1.2, for any integer
m ≥ 1 the germ of this curve has the form
x = t, y = t(1 + a₁t + … + aₘtᵐ + O(tᵐ⁺¹)) (6.8)
with some constants a₁, …, aₘ. We break the further proof
into several steps.
STEP 1: KILLING THE MONOMIAL t². Without loss of generality we may
assume a₁ = 1, since any a₁ ≠ 0 can be reduced to 1 by multiplying the parameter t and the coordinates x, y by suitable nonzero numbers. It therefore suffices to prove the required statement for
the curve x = t³, y = t + t⁵ + O(t⁶). The substitution x ↦ x + 3y turns this
curve into
x = φ(t) = t + a(t + t²) + O(t³), y = y(t) = t + t² + O(t³).
Now let us make the change of parameter t ↦ t̃ by the formula t̃ = φ(t), i.e.
t̃ = t(1 + a(t + t²) + O(t³)) → t = t̃ — a·t̃² + O(t̃³).
After this our curve takes the form
x = t̃, y = t̃(1 + a·t̃ + O(t̃²)) + t̃² — a·t̃² + O(t̃³)) = t̃ + O(t̃⁶).
STEP 2: KILLING THE REMAINING MONOMIALS. Consider the curve
x = t, y = t(1 + a₄t⁴ + … + aₘtᵐ + O(tᵐ⁺¹)), (6.9)
and let us show that, using a suitable change of the variables (x, y), given by a formal power series, one can kill all monomials tᵏ
of degree k ≥ 5. It suffices to note that for any k ≥ 5 the equality
x = tⁱyʲ + O(tᴺ) holds with some integers i, j ≥ 0. For example,
t⁵ = x², t⁶ = x²y + O(t⁷), t⁷ = xy + O(t⁸),
t⁸ — x³, t⁹ — x² + O(t¹⁰), t¹⁰ — xy + O(t¹¹),
… tᵏ = …, tᵏ⁺¹ = x²y + O(tᵏ⁺²), tᵏ⁺² = y + O(tᵏ⁺³), …
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Consequently, using the formal substitution
∞
y ↦ y + Σ cᵢⱼ xⁱyʲ (6.10) i+j ≥2
with suitable coefficients cᵢⱼ, we can kill all the monomials tᵏ, k ≥ 6, in (6.9). Namely, the monomial cᵢⱼxⁱyʲ kills the monomial tᵏ, where i+j+1 = k,
and does not change the coefficients of monomials of degree less than k.
If the power series on the right-hand side of formula (6.10) converges in some neighborhood of the origin, it defines there an
analytic function f(x,y), and the substitution y ↦ y + f(x, y) completes the proof of our statement. However, in general, series (6.10)
may have zero radius of convergence, i.e. turn out to be merely formal.
To overcome this difficulty we take the following step.
STEP 3: KILLING THE INFINITELY FLAT REMAINDER. By
Lemma 5.1 of Section 5.3, there exists a smooth function f(x,y), whose Taylor series
at the origin coincides with the formal series
from formula (6.10). Then, clearly, the substitution y ↦ y + f(x,y) brings
our curve to the form
x = t, y = t — h(t), (6.11)
where the function h is smooth and infinitely flat at zero. It is easy to verify that the function g(x) = h(x^{1/3}) is likewise smooth
and infinitely flat at zero. (This can be done, for example, using representation (1.5) of Section 1.1.) Then, clearly, the substitution
y ↦ y — g(x) brings the germ of curve (6.11) to the desired form. |
We shall call the germs of two curves equivalent if one
of them is taken to the other by a smooth change of the parameter t
and a smooth change of the variables (x, y). For example, as we saw above,
the germ of any curve (6.6) with exponent n = 2 or 3 is brought to
the normal form (6.3). This might suggest that the germ of
curve (6.6) with any n is brought to the normal form (6.3); however,
for n ≥ 4 this is no longer true. Note, however, that in Section 8 another,
weaker equivalence relation will be
introduced, under which the germ of curve (6.6) with any n is equivalent to (6.3).
PROBLEM 6.5. Prove that, by means of a smooth change of the parameter t and a smooth change of the variables (x, y), the germ of curve (6.6) with exponent n = 4 is brought to one of three mutually inequivalent
normal forms:
x = t, y = t⁴; x = t, y = t⁴ + t⁵. (6.12)
Up to now we have used substitutions (i.e. local diffeomorphisms) of class C∞. It is natural to ask: what happens
if we relax the smoothness requirement to a finite one? We shall call two
germs C^k-equivalent if they can be turned into each other by a C^k-smooth change of the parameter t and a C^k-smooth change of the variables (x, y). Then some germs that are not C∞-
equivalent may turn out to be C^k-equivalent for some
integer k ≥ 1. The simplest example:
PROBLEM 6.6. Prove that the germ of curve (6.6) with any n is C¹-equivalent to the germ (6.3) with the same exponent n.
SOLUTION. In the case of odd exponent n it is convenient to use formula (6.6) with φ(t) = t, ψ(t) = 1 + g(t), g(0) = 0 (see Problem 6.2).
Then, using the expansion g(t) = g₁t + … + gₙ₋₁tⁿ⁻¹ + gₙtⁿ,
we obtain
x = t, y = t(1 + g(t)) = tⁿ(1 + … + gₙtⁿ).
It follows that the C¹-smooth change of variables
y ↦ (a₁tⁿ + … + gₙ₋₁tⁿ⁻¹)
— 1 — 1+2 (22)
brings our curve to the normal form x = tⁿ, y = tⁿ⁺¹.
In the case of even n it is convenient for the proof to use
formula (6.6) with the functions φ(t) = 1 + g(t) and ψ(t) = 1. All further arguments are analogous. The reader is also invited to answer independently the question: can germs (6.3) with different n
be C¹-equivalent? |
PROBLEM 6.7. Prove that the normal forms (6.12) are not C²-equivalent.
REMARK 6.1. The question of the equivalence of germs of curves
(6.6), and even of the more general form (6.5), has been touched upon in many works;
let us mention only the papers [4, 12, 10, 19] and the book [41]. Here the most studied case has been analytic equivalence of germs, in which
both the functions φ(t), ψ(t) defining the curve, and the change of parameter t, and the change of
variables (x, y) are assumed to be analytic (in the real
or complex domain). As far as we know, finite-smooth
equivalence has hardly been studied at all.
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This question seems interesting even if one restricts oneself to
curves of the form (6.6). As we have just seen, all germs (6.6) with a given exponent n are C¹-equivalent to one another. On the other hand,
it follows from the results of [12] that for any exponent n ≥ 5
there exist infinitely many germs of curves of the form (6.6) that are not analytically equivalent. It is not known how matters stand between these
two «extreme cases». Perhaps this is an interesting and nontrivial problem awaiting its researcher.
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