Frankel 2.4 Problems


The second rank tensor given by \(a_ib_j dx^i \otimes dx^j\) has the value on vectors \(v\) and \(w\):

$$a_ib_j dx^i(v) \otimes dx^j(w) = a_iv^i b_jw^j.$$

For one-forms \(\alpha\) and \(\beta\) with components \(a_i\) and \(b_j\), the tensor product \(\alpha \otimes \beta\) has the value on vectors \(v\) and \(w\):

$$\alpha(v) \otimes \beta(w) = a_iv^i b_jw^j.$$


$$\alpha \otimes \beta = a_ib_j dx^i \otimes dx^j.$$



let \(\mathbf{A} : E \to E\) be a linear transformation.

(i) The trace tr(\(\mathbf{A}) = A_i^i\). If we change coordinate systems from \(x\) to \(y\) the components of the mixed tensor of \(\mathbf{A}\) transform as

$$A'^l_k = \frac{\partial x^j}{\partial y^k}\frac{\partial y^l}{\partial x^i}A^i_j.$$

Therefore, tr(\(\mathbf{A})\) is a scalar:

$$A'^k_k = \frac{\partial x^i}{\partial y^k}\frac{\partial y^k}{\partial x^i}A^i_i = A^i_i,$$



Let \(\vec{v} = v^i\partial_i\) be a vector on \(M^n\).

(i) We define a new object by \(v_j = g_{ji}v^i\). By the transformation properties,

$$v'_k = \frac{\partial x^i}{\partial y^k}\frac{\partial x^j}{\partial y^k}g_{ji}\frac{\partial y^k}{\partial x_i}v^i = \frac{\partial x^j}{\partial y^k}g_{ji}v^i = \frac{\partial x^j}{\partial y^k}v_j,$$

thus, \(v_j\) defines a one-form.


Frankel 2.3 Problems


Let \(F : M^n \to W^r\) and \(G : W^r \to V^s\) be smooth maps Let \(x\), \(y\), and \(z\) be local coordinates near \(p \in M^n\), \( F(p) \in W^r\), and \( G(F(p)) \in V^s\), respectively. Consider the composite map \(G \circ F : M \to V\),

(i)     Using the coordinate basis \(\partial / \partial x\), we may investigate the differential \((G \circ F)_*\).

$$(G \circ F)_*\left(\frac{\partial}{\partial x^i}\right) = \sum_{k=1}^s \frac{\partial z^k}{\partial x^i} \frac{\partial}{\partial z^k}.$$

By the chain rule,

$$(G \circ F)_*\left(\frac{\partial}{\partial x^i}\right) = \sum_{k=1}^s \sum_{j=1}^r \frac{\partial z^k}{\partial y^j}\frac{\partial y^j}{\partial x^i}\frac{\partial}{\partial z^k}.$$


$$F_*\left(\frac{\partial}{\partial x^i}\right) = \sum_{j=1}^r \frac{\partial y^j}{\partial x^i} \frac{\partial}{\partial y^j},$$

we see that 

$$(G \circ F)_*\left(\frac{\partial}{\partial x^i}\right) = \sum_{k=1}^s \sum_{j=1}^r F_*^j\left(\frac{\partial}{\partial x^i}\right)\frac{\partial z^k}{\partial y^j}\frac{\partial}{\partial z^k},$$

and conclude

$$(G \circ F)_* = G_* \circ F_*.$$


(ii)   Using the covector coordinate basis \(dx\), we may investigate the pull-back \((G \circ F)^*\).

$$(G \circ F)^*(dz^k) = \sum_{i=1}^n \frac{\partial z^k}{\partial x^i} dx^i.$$

By the chain rule,

$$(G \circ F)^*(dz^k) = \sum_{i=1}^n \sum_{j=1}^r \frac{\partial z^k}{\partial y^j} \frac{\partial y^j}{\partial x^i} dx^i.$$


$$G^*(dz^k) = \sum_{j=1}^r \frac{\partial z^k}{\partial y^j} dy^j,$$

we see that

$$(G \circ F)^*(dz^k) = \sum_{i=1}^n \sum_{j=1}^r G_j^*(dz^k) \frac{\partial y^j}{\partial x^i} dx^i.$$

and conclude that

$$(G \circ F)^* = F^* \circ G^*.$$


Frankel 2.1 Problems


If \(\mathbf{v}\) is a vector and \(\alpha\) is a covector, then

\sum a_i^V v_V^i &= \sum_i \left[ \sum_j \frac{\partial x_U^j}{\partial x_V^i}a_j^U \right] \left[ \sum_k \frac{\partial x_V^i}{\partial x_U^k}v_U^k \right] \\
&= \sum_{i,j,k} \frac{\partial x_U^j}{\partial x_V^i} \frac{\partial x_V^i}{\partial x_U^k}a_j^U v_U^k.


$$\frac{\partial x_U^j}{\partial x_V^i} \frac{\partial x_V^i}{\partial x_U^k} = \delta_k^j,$$

we have

$$\sum a_i^V v_V^i = \sum a_i^U v_U^i.$$

If instead we were to look at the sum \(\sum v^i w^i\) where \(\mathbf{w}\) is another vector, then we would get a coordinate dependent result.



Let \(x^1\), \(x^2\), and \(x^3\) be the usual cartesian coordinates in \(\mathbb{R}^3\) and let \(u^1 = r\), \(u^2 = \theta\) (colatitude), and \(u^3 = \phi\) be spherical coordinates.

(i)    The metric tensor for spherical coordinates is defined

$$g_{ij} = \langle \partial_{i}, \partial_{j} \rangle.$$

Since we know \(g_{ij} = \delta_{ij}\) in cartesian coordinates, we can easily get the metric in spherical coordinates by the transformation

$$g_{ij} = \sum_{rs} \frac{\partial x^r}{\partial u^i}\frac{\partial x^s}{\partial u^j}\delta_{rs}.$$

Therefore the metric tensor components are

g_{11} &= 1, \\
g_{22} &= r^2, \\
g_{33} &= 2r^2\cos^2\theta, \\
g_{12} &= g_{21} = 0, \\
g_{13} &= g_{31} = 0, \\
g_{23} &= g_{32} = 0.

(ii)   We can use this to compute the components of the gradient vector \(\nabla f\),

$$(\nabla f)^i = \sum_j g^{ij}\frac{\partial f}{\partial u^j}.$$

So then we have

(\nabla f)^1 &= 1, \\
(\nabla f)^2 &= 1/r^2, \\
(\nabla f)^3 &= 1/2r^2\cos^2\theta.\\

(iii)   The orthogonality of the spherical basis is evident from the diagonality of \(g\), and the non-normality is evident from the diagonal itself. If we rescale the coordinate system we can make it orthonormal. To do this we divide every coordinate vector by its magnitude. This makes the gradient 

$$\nabla f = \frac{\partial f}{\partial r} + \frac{1}{r}\frac{\partial f}{\partial \theta} + \frac{1}{2r\cos\theta}\frac{\partial f}{\partial \phi}.$$


Frankel 1.3 Problems


What would be wrong in defining

$$||X||^2 = \sum_i (X_U^i)^2 \,\,\, ?$$

Defining the magnitude this way would lead to a result that depends on the choice of coordinates.


Let the torus \(T^2\) be embedded in \(\mathbb{R}^3\) such that it lies flat atop the xy plane. The map \(F : T^2 \to \mathbb{R}^2\) project points on the torus to the xy plane.  By inspection, the critical points of \(F\) are those along the inner and outer circles in the middle of the torus. At these points the tangent space is perpendicular to the xy plane and therefore the differential \(F_*\) is not onto.

Frankel 1.1 Problems


The locus described by \(x^2 + y^2 - z^2 = c\) is a 2 dimensional submanifold of \(\mathbb{R}^3\) for \(c < 0\) and \(c = 0\) but not for \(c > 0\). This can be seen by investigating the functional dependence

$$z = z(x,y) = \sqrt{x^2+y^2-c}.$$

For \(c < 0, z(x,y)\) is differentiable over it's entire domain, by definition forming a 2D submanifold of \(\mathbb{R}^3\). For \(c = 0, z(x,y)\) is differentiable everywhere except at \(x = y = 0\), and therefore forms a submanifold if and only if the origin is omitted. For \(c > 0, z(x,y)\) is defined on the reals only for \(x^2+y^2 \ge c\) and differentiable only for \(x^2+y^2 > c\), so it only forms a submanifold for \(x^2+y^2 > c\).



SO(\(n\)) is defined to be the set of all orthogonal \(n \times n\) matrices \(x\) with \(\det x = 1\). The discussion on SO(3) in the book easily generalizes to show that SO(\(n\)) is an \(n\)-dimensional submanifold of \(\mathbb{R}^{n^2}\).



The special linear group is the set

$$\mathrm{SL}(n) := \{ n\times n \text{ real matrices } x | \det x = 1 \}.$$

Clearly SL(\(n\)) is a subset of the set of \(n \times n\) matrices, which itself is isomorphic to \(\mathbb{R}^{n^2}\). The condition on the determinant restricts SL(\(n\)) to be a space of dimension \(n^2 - 1\), but to show it's a submanifold we must demonstrate that \(\det x = 1\) defines a differentiable map. To do this we first expand the determinant around the \(j\)'th row,

$$\det x = \sum_{i=1}^n (-1)^{i+j} M_{ij}x_{ij}.$$

Here, \(M_{ij}\) is the minor.  Taking the derivative with respect to \(x_{kj}\),

$$\frac{\partial}{\partial x_{kj}}\det x = \sum_{i=1}^n (-1)^{i+j}M_{ij}\frac{\partial x_{ij}}{\partial x_{kj}} = (-1)^{k+j}M_{kj}.$$

As long a \(M_{kj}\) is non-zero, the implcit function theorem tells us that \(x_{kj}\) can be expressed as a differentiable function of the other components. Therefore, by definition, SL(\(n\)) is a \(n^2 - 1\) submanifold of \(\mathbb{R}^{n^2}\).

Frankel 1.2 Problems


The real projective plane \(\mathbb{R}P^2\) is defined the be the set of all lines in \(\mathbb{R}^3\) that pass through the origin. We must show that this forms a 2-dimensional manifold. We can cover this set with three subsets:

$$\begin{align}U_x &:= \text{all lines which do not lie on the yz-plane}, \\
U_ y &:= \text{all lines which do not lie on the xz-plane}, \\
U_z &:= \text{all lines which do not lie on the xy-plane}.\end{align}$$

For each of these we can define a coordinate system by associating each line is the set with a point in the plane. That is, a map \(\phi_i(U_i) :  \mathbb{R}P^2 \to \mathbb{R}^2, \text { for } i = x,y,z\). To do this for \(U_z\), chose any point \((x,y,z)\) other than the origin on the line and construct

$$\phi_z(U_z) = \left(\frac{x}{z},\frac{y}{z}\right).$$

So a line in \(U_z\) gets mapped to a point in \(\mathbb{R}^2\) which is the intersection of that line and the \(z=1\) plane. The other coordinate patches are defined similarity. 

To show that \(\mathbb{R}P^2\)  forms a manifold, we must show that for any line existing in more than one coordinate patch, the coordinates in one patch can be differentiably expressed in terms of the coordinates in another patch.

$$\phi_x(\phi_z^{-1}(a,b)) = \phi_x(x,y,1) = \left(\frac{y}{x},\frac{1}{x}\right).$$

This is \(C^\infty\) over \(U_z \cap U_x\) (\(x \neq 0\) in \(U_x\)). Likewise for all other patch relationships. Therefore, the real projective plane is a 2-dimensional manifold.



The real projective space \(\mathbb{R}P^3\) is a 3-dimensional manifold. A point \(p = (x^1,x^2,x^3,x^4)\) is identified with \((\lambda x^1, \lambda x^2, \lambda x^3,\lambda x^4)\) for all \(\lambda \neq 0 \in \mathbb{R}\). Following the example of \(\mathbb{R}P^2\), we cover \(\mathbb{R}P^3\) with the 4 coordinate patches

$$U_i = \{(\lambda x^1, \lambda x^2, \lambda x^3,\lambda x^4) \, | \, x^i \neq 0\}, \,\,\, i = 1,2,3,4.$$

Then the coordinate maps \(\phi_i : U_i \to \mathbb{R}^3\) are clear, e.g.

$$\phi_1(U_1) = \left(\frac{x^2}{x^1}, \frac{x^3}{x^1}, \frac{x^4}{x^1}\right).$$


$$\phi_2(\phi_1^{-1}(a,b,c,d)) = \phi_2(1,x^2,x^3,x^4) = \left(\frac{1}{x^2}, \frac{x^3}{x^2}, \frac{x^4}{x^2}\right),$$

which is \(C^\infty\). Hence \(\mathbb{R}P^3\) is a 3-dimensional manifold.