# FAQ

##### What does $$\mathcal{E}_n^2$$ mean?
• It's defined to be
$$\mathcal{E}_n^2 = \left\{\begin{pmatrix} f \cr g\end{pmatrix} \mid f,g \in \mathcal{E}_n\right\}$$
That is $$\mathcal{E}_n^2 \simeq \mathcal{E}_n \times \mathcal{E}_n$$. Similarly $$\mathcal{E}_n^p$$ is the Cartesian product of p copies of $$\mathcal{E}_n$$. Note that if $$f:(\mathbb{R}^n,0)\to(\mathbb{R}^p,0)$$ then $$\theta(f) = \mathcal{E}_n^p$$.
##### How are the $$\mathcal{R}$$- and $$\mathcal{K}$$- finite determinacy theorems related

In both cases, the condition is $\mathfrak{m}_n^{k+1}\theta(f) \ \subset \ \mathfrak{m}_n\,T\mathcal{G}\cdot f.$

For $$\mathcal{R}$$-equivalence, $$T\mathcal{R}\cdot f = \mathfrak{m}_n\,Jf$$, and recall that with p=1, $$\theta(f) \equiv \mathcal{E}_n$$, so the left-hand-side is just $$\mathfrak{m}_n^{k+1}$$.

##### Problem sheet 2

You will have noticed that some of the solutions are missing. Those problems are for general understanding, and not specifically necessary for the syllabus.

##### Tangent space $$T\mathcal{K}\cdot f$$

Note that what I called $$Jf$$ in the lectures is the same as $$tf(\theta(n))$$ and is also denoted $$T\mathcal{R}_e\cdot f$$.

Similarly, $$\mathfrak{m}_n\,Jf = tf(\mathfrak{m}_n\theta(n)) = T\mathcal{R}\cdot f$$.

The example I did in lectures $$f(x,y) = (x^2,\,y^2)$$ is also done in Chapter 11 (see p.114). For another example, see here.

#### I'm just not too sure what 'parametrizing $$C_F$$ by (x,y,u)' means. Do we only keep the (x,y,u) terms?

Remember (1st year?) spherical polar coordinates: varying $$\theta$$ and $$\phi$$ give different points on the sphere (r = contant). That is a parametrization of the sphere by $$\theta$$ and $$\phi$$ (one has $$x= \sin\theta\, \cos\phi$$ etc). Similarly, here if we vary x,y,u, by putting v = (function of x,y,u) (much like $$x=\sin \theta$$ etc) we get different points on $$C_F$$. So that is a parametrization of $$C_F$$.

If one needs to do any calculations on the sphere or on $$C_F$$, it is enough to do them using the parameters: eg finding the points where $$\pi_F$$ is not a local diffeo - just write $$\pi_F$$ in terms of $$x,y,u$$ (and find its Jacobian etc).

#### What is a cobasis?

Definition A cobasis of I in $$\mathcal{E}_n$$ is a set of functions (usually chosen to be monomials) $$h_1,h_2,\dots,h_k$$ such that given any function $$f\in\mathcal{E}_n$$ there are unique real numbers (coefficients) $$a_1,\dots,a_k$$ and a function $$g\in I$$ such that $f = a_1h_1+\cdots + a_kh_k \ + \ g.$ (The uniqueness of the coefficients is what makes it a cobasis - otherwise they are just generators for $$\mathcal{E}_n/I$$, see properties of basis in linear algebra).

If we want the cobasis for I in $$\mathfrak{m}_n$$ instead, then replace $$\mathcal{E}_n$$ by $$\mathfrak{m}_n$$ in the definition above (in practice the difference is whether we include 1 in the cobasis or not).

Example Consider $$I=\left<xy,\,x^3+y^3\right> \subset \mathcal{E}_2$$ (see Ex 3.9(ii)). There is a choice of cobasis. For example $$\{1,x,y,x^2,y^2,x^3\}$$. Another is $$\{1,x,y,x^2,y^2,y^3\}$$. But you can't have both $$x^3$$ and $$y^3$$, as if you did the coefficients would not be unique. For example, if you take $$f=2x^3+3y^2$$, and you allow both you can write $f = 2x^3+3y^2 + (0)\quad\mbox{as well as}\quad f = 3y^2-2y^3 + (2x^3+2y^3)$ (The term in brackets ( ) is the $$g\in I$$.) More examples can be found in Exercise 3.9.

Further discussion: Consider the quotient $$\mathcal{E}_n/I$$ (which is a vector space). A cobasis for an ideal $$I$$ in $$\mathcal{E}_n$$ is essentially a basis for this quotient. However, strictly speaking, the elements of $$\mathcal{E}_n/I$$ are not functions but are of the form $$f+I$$ where $$f\in\mathcal{E}_n$$, rather than just $$f$$; one says $$f$$ is a representative of $$f+I$$. Then if $$h_1,h_2,\dots,h_k$$ are representatives such that $\{h_1+I,\,h_2+I,\,\dots,\,h_k+I\}$ is a basis for $$\mathcal{E}_n/I$$, then $$\{h_1,h_2,\dots,h_k\}$$ is a cobasis of $$I$$ in $$\mathcal{E}_n$$.

#### Calculations with ideals (07/11/09)

##### Why is $$\left<2x^2,\,3y^2\right> = \left<x^2,\,y^2\right>$$, say?
The answer is simply that since 1/2 is in the ring of germs $$\mathcal{E}_2$$ (it's a constant function), if $$2x^2\in I$$ then it follows that $$(1/2)2x^2\in I$$ (definition of ideal: $$h\in I, k\in R \Rightarrow kh\in I$$). Similarly, $$y^2 = (1/3)(3y^2)$$.
On the other hand, one cannot simplify say $$\left<x+3y^2,\,xy^3\right>$$ - it is not equal to $$\left<x+y^2,\,xy^3\right>$$.
##### How do we choose a basis for $$\mathcal{E}_2/Jf$$ when Jf is not generated by monomials?
For example, if $$Jf = \left<x^2+y^2,\,xy\right>$$ then the monomials in the ideal are $$x^3,\, xy,\, y^3$$ (I hope you agree). The monomials not in $$Jf$$ are therefore $$1,\,x,\, y,\,x^2,\,y^2$$.. However, the sum of the last two is in $$Jf$$, so they are not both needed - but you must include one of them, so one basis for $$\mathfrak{m}/Jf$$ is $$\{x,\,y,\,x^2\}$$. Or instead of $$x^2$$ we could use $$y^2$$, or $$x^2-y^2$$, or any other linear combination of the two except (multiples of) $$x^2+y^2$$.