# Computing pullback

In my differential geometry course, we were given the following example, but, the instructor did not explain how he got the answer, so I am stuck. It was:

Let $\phi: \mathbb{R}^2 \to \mathbb{R}^3$ be defined by:

$\phi(u,v) = (u, v, u^2 + v^2)$, and let $g_0$ be the Euclidean metric tensor on $\mathbb{R}^3$. Then:

$g_1 = \phi^{*}g_0 = (1 + 4u^2)du^2 + (4uv)du dv + (4uv)dvdu + (1+4v^2)dv^2$.

I don't understand how he got this result. I understand from a theorem previously introduced in lecture that:

$(\phi^{*}g)(X_p, X_p) = g(\phi_{*}X_p, \phi_{*}X_{p})$, where $X_p$ is a vector field in the tangent space at $p$, and $\phi_{*}$ is the Jacobian matrix, derivative map of $\phi$. I don't know what to do once you get this derivative matrix.

Thanks!

## 1 Answer

$\phi^* dx = du$, $\phi^* dy = dv$, and $\phi^*dz = 2(u\,du+v\,dv)$, so \begin{align*} \phi^*(dx\otimes dx + dy\otimes dy + dz\otimes dz) &= du\otimes du + dv\otimes dv + 4(u\,du+v\,dv)\otimes (u\,du+v\,dv) \\ &= (1+4u^2)du\otimes du + 4uv(du\otimes dv + dv\otimes du) + (1+4v^2)dv\otimes dv. \end{align*} Your instructor is suppressing the tensor signs, as was standard in classic treatments.

• Hi. Thank you that is very helpful! Just a quick question, does the "middle term" cross-term cancel out. I thought tensor products are anti-symmetric, so, $du \wedge dv = -dv \wedge du$? – Thomas Moore Oct 9 '17 at 22:49
• Noooo ... Wedge products (appearing in differential forms) are skew-symmetric. The metric is a symmetric tensor ... which is why the matrix for the first fundamental form is always symmetric. :) – Ted Shifrin Oct 9 '17 at 22:51