T Kf

Another Example of \(\mathcal{K}\)-tangent space(s) and cobases: Let \(f(x,y)= ( x^2+y^2,\, y^3)\).

We have \[Jf\equiv tf(\theta(2)) = \mathcal{E}_2\left\{\pmatrix{2x\cr 0},\; \pmatrix{2y\cr 3y^2}\right\}\]

 and 

\[ I_f\theta(f) = \mathcal{E}_2\left\{\pmatrix{x^2+y^2\cr 0},\; \pmatrix{y^3\cr 0},\; \pmatrix{0\cr x^2+y^2},\; \pmatrix{0\cr y^3}\right\}.\]

Thus \[T\mathcal{K}_e\cdot f = \mathcal{E}_2\left\{ \pmatrix{x\cr 0},\; \pmatrix{2y\cr 3y^2},\; \pmatrix{x^2+y^2\cr 0},\; \pmatrix{y^3\cr 0},\; \pmatrix{0\cr x^2+y^2},\; \pmatrix{0\cr y^3}\right\}.\] while \[T\mathcal{K}\cdot f = \mathcal{E}_2\left\{ \pmatrix{x^2\cr 0},\; \pmatrix{xy\cr 0},\; \pmatrix{2xy\cr 3xy^2},\; \pmatrix{2y^2\cr 3y^3},\; \pmatrix{x^2+y^2\cr 0},\; \pmatrix{y^3\cr 0},\; \pmatrix{0\cr x^2+y^2},\; \pmatrix{0\cr y^3}\right\}.\] (The first four terms come from \(\mathfrak{m}_2Jf = tf(\mathfrak{m}_2\theta(2))\), so from multiplying each of the generators of \(Jf\) above by each of \(x\) and \(y\) in turn.) Call these 8 generators \(\alpha,\beta,\gamma,\delta,\epsilon,\zeta,\eta,\theta\) in turn.

First consider the monomials of the from \(\pmatrix{x^ay^b\cr 0}\). We have, \(\pmatrix{x^2\cr0}=\alpha,\;\pmatrix{xy\cr0}=\beta,\;\pmatrix{y^2\cr0} = \epsilon-\alpha\) so giving all of \(\pmatrix{\mathfrak{m}_2^2\cr0}\) (ie all of \(\mathfrak{m}_2^2\) in the top row).

Now consider the monomials of the from \(\pmatrix{0\cr x^ay^b}\). there are no quadratic monomials - only \(\eta\) which is not monomial. However, there are plenty of cubics: \[\pmatrix{0\cr y^3} = \theta,\quad \pmatrix{0\cr x^2y} = y\eta-\theta,\quad \pmatrix{0\cr xy^2}=\frac13(\gamma-2\beta),\quad \pmatrix{0\cr x^3} = x\eta -\frac13(\gamma-2\beta). \] Thus we can take the following as a cobasis of \(T\mathcal{K}\cdot f\) in \(\mathfrak{m}_2\theta(f)\): \[\left\{ \pmatrix{x\cr 0},\;\pmatrix{y\cr 0},\; \pmatrix{0\cr x},\; \pmatrix{0\cr y},\; \pmatrix{0\cr x^2},\;\pmatrix{0\cr xy}\right\}.\] The \(\mathcal{K}\)-codimension is therefore equal to 6. [Note that we don't need both \(\pmatrix{0\cr x^2}\) and \(\pmatrix{0\cr y^2}\) because their sum is in \(T\mathcal{K}\cdot f\).]

For a cobasis of \(T\mathcal{K}_e\cdot f\) in \(\theta(f)\), we can take (see \(T\mathcal{K}_e\cdot f\) above), \[\left\{\pmatrix{1\cr0},\; \pmatrix{y\cr0},\; \pmatrix{0\cr 1},\; \pmatrix{0\cr x},\; \pmatrix{0\cr y},\; \pmatrix{0\cr xy}\right\}.\]