Welcome to the Math Page.
This site is where I go through topics in differential geometry following Barret O'Neill
"Elementary Differential Geometry".
$$ {J(\theta) =\frac{1}{2m} \sum^m_{i=1}(h_\theta(x^{(i)}) - y^{(i)})2 + \lambda\sum^n_{j=1}\theta^2_j} $$
Differential Geometry is a branch of mathematics that lies at the heart of all motion in physics. It incorporates everything that the physicist
needs to describe nature in a quantitative manner, albeit classically. However, work has been done to try and use differential geometry in quantum mechanics.
And, what's more is that the fundamental concepts in differential geometry relating to the first fundamental form, curvature, invariants of curves can
all be related to fundamental principles in physics. Another reason is that differential geometry is really the gateway to understanding General Relativity beyond
the conceptual ideas such Lorentz contraction and time dilation.
For those reasons alone, differential geometry is a crucial part of the theoretical physicist's tool kit.
We start with defining points in \(\mathbb{R}^3 \). These are just what you might three numbers, \( \boldsymbol{p}=(p_1, p_2, p_3) \), that describe a location in \(\mathbb{R}^3 \). You can add any two points \[ \boldsymbol{a}+\boldsymbol{b}=(a_1+b_1,a_2+b_2,a_3+b_3) \] You can also multiply by a scalar, a real number, \( c \in \mathbb{R} \). \[ c\boldsymbol{p}=(cp_1, cp_2, cp_3) \] This space of points is isomorphic to what we usually would call a three dimensional vector space, but we'll follow O'Neill and be careful saying what is a vector and what is a point because soon we will have the notion of a vector being described by two connecting points.
Now, that we have points in \( \mathbb{R}^3\), we can now say that the points give us the coordinates. In first semesters of Calculus, these are the usual \((x,y,z) \) coordinates. But, now we will consider them as functions themselves. So that \[ \boldsymbol{p}=(x( \boldsymbol{p}), y( \boldsymbol{p}), z( \boldsymbol{p})) \] and we see that \( x, y, z\) are considered functions of the points in \( \mathbb{R}^3\) space. The author makes the point that when first learning calculus that no real distinction is made between points and coordinate functions because \[ x( \boldsymbol{p}) = x=x_1, y( \boldsymbol{p})=y=x_2, z( \boldsymbol{p})=z=x_3 \] We define the notation \( x = x_1, y = x_2, z = x_3 \) since it will be used later on for general coordinate functions. This means then the coordinate functions are themselves functions of points and do not have to be a simple relation like the one with which we are very accustomed.
A couple properties for these functions - they can be added point-wise and we will assume that they are all at least twice differentiable.
Being point-wise means that if we have two functions such as \( x(\boldsymbol{p}) + y(\boldsymbol{p}) \), we can add the values of the functions for the same point, \( \boldsymbol{p}\). \[ (x+y)(\boldsymbol{p}) = x(\boldsymbol{p}) + y(\boldsymbol{p}) \] Multiplication works similarly, \[ (xy)(\boldsymbol{p}) = x(\boldsymbol{p})y(\boldsymbol{p}) \] This idea will be used a lot including using vector fields as functions.
On the idea that the functions need to twice differentiable just means that a function must not end up being zero after the second derivative. Although not necessarily on the number of times a function is differentiable, there is also the idea that differentiation is also a local operation, not global. The argument is that when differentiating a function, it is only necessary to know points in the neighborhood of the derivative near, \( \partial f / \partial x \) that point, \( \boldsymbol{p} \). Then, domain of \(f \) only needs to be an open set \( \mathcal{O} \) where \( \boldsymbol{p} \in \mathcal{O} \).
1. Let \( f=x^2 y \) and \( g=y\sin z \) be functions on \(\mathbb{R}^3 \). Express the following functions in terms of \(x, y, z\):
1.a. \[ fg^2 = (x^2 y )(y\sin z)^2 =x^2 y^3 \sin^2 z \]
1.b. \[ \frac{\partial f}{\partial x} g + \frac{\partial g}{\partial x} f = \frac{\partial x^2 y}{\partial x} y\sin z + \frac{\partial y\sin z}{\partial y} x^2 y \] \[ = \frac{\partial x^2 y}{\partial x} y\sin z + \frac{\partial y\sin z}{\partial y} x^2 y = 2xy^2\sin z + x^2 y \sin z \]
1.c. \[ \frac{\partial^2 (fg)}{\partial y \partial z} = \frac{\partial }{\partial y} \frac{\partial }{\partial z} x^2 y^2 \sin z = \frac{\partial }{\partial y} x^2 y^2 \frac{\partial }{\partial z} \sin z = \cos z \frac{\partial }{\partial y} x^2 y^2 \] \[ = 2x^2 y \cos z \]
1.d. \[ \frac{\partial}{\partial y} (\sin f) = \frac{\partial}{\partial y} (\sin x^2y)\] \[ = \frac{\partial}{\partial y} (\sin x^2y) = \frac{\partial \sin x^2y}{\partial (x^2y)} \frac{\partial x^2y}{\partial y} = x^2 \cos x^2y \]
2. Find the value of the function \( f=x^2y-y^2z \) at each point:
2.a. \[ f=x^2y-y^2z \] Note that \[ \boldsymbol{p}=(x( \boldsymbol{p}), y( \boldsymbol{p}), z( \boldsymbol{p})) =(p_1=1, p_2=1,p_3=1 ) \] Then, \[ f=x^2y-y^2z =(1)^2(1)-(1)^2(1)=0 \]
2.b. \[ f=x^2y-y^2z \] Note that \[ \boldsymbol{p}=(x( \boldsymbol{p}), y( \boldsymbol{p}), z( \boldsymbol{p})) =(3, -1,1/2 ) \] Then, \[ f=x^2y-y^2z =(3)^2(-1)-(-1)^2(1/2)=-9-\frac{1}{2} = 8\frac{1}{2} \]
2.c. \[ f=x^2y-y^2z \] Note that \[ \boldsymbol{p}=(x( \boldsymbol{p}), y( \boldsymbol{p}), z( \boldsymbol{p})) =(a,1,1-a) \] Then, \[ f=x^2y-y^2z =(a)^2(1)-(1)^2(1-a)= a^2 -1+a=a^2+a-1 \]
2.d. \[ f=x^2y-y^2z \] Note that \[ \boldsymbol{p}=(x( \boldsymbol{p}), y( \boldsymbol{p}), z( \boldsymbol{p})) = ( t, t^2, t^3 ) \] Then, \[ f=x^2y-y^2z =(t)^2(t^2)-(t^2)^2(t^3)= t^4 -t^7 \]
3. Express \( \partial f / \partial x \) in terms of \( x,y,z \) if
3.a. \[ \frac{\partial }{\partial x }f= \frac{\partial }{\partial x } ( x\sin (xy) +y\cos (xz) ) \] \[ = \frac{\partial }{\partial x } x\sin (xy) + \frac{\partial }{\partial x } y\cos (xz) \] \[ = \frac{\partial x}{\partial x } \sin (xy) + \frac{\partial \sin (xy)}{\partial x } x + \frac{\partial y }{\partial x } \cos (xz) + \frac{\partial \cos (xz) }{\partial x } y\] \[ = \sin (xy) + xy\cos (xy) + 0 + - yz \sin (xz) \] \[ = \sin (xy) + xy\cos (xy) - yz \sin (xz) \]
3.b. \[ \frac{\partial f }{\partial x }= \frac{\partial \sin g }{\partial x } \] Also note, \( f=\sin g, g=e^h, h=x^2+y^2+z^2 \), so we have a chain rule here. \[ \frac{\partial f }{\partial x }= \frac{\partial f(g) }{\partial g }\frac{\partial g(h) }{\partial h } \frac{\partial h(x) }{\partial x }\] \[ \frac{\partial f(g) }{\partial g } = \frac{\partial \sin g }{\partial g } = \cos(g) \] \[ \frac{\partial g(h) }{\partial h } = \frac{\partial e^h }{\partial h } = e^h \] \[ \frac{\partial h(x) }{\partial x } = \frac{\partial (x^2+y^2+z^2) }{\partial x } = 2 x \] Then, \[ \frac{\partial f }{\partial x }= \frac{\partial f(g) }{\partial g }\frac{\partial g(h) }{\partial h } \frac{\partial h(x) }{\partial x } = \cos(g) (e^h) (2 x) = 2 x e^{x^2+y^2+z^2} \cos(e^{x^2+y^2+z^2}) \]
4. If \( g_1,g_2,g_3 \) and \( h \) are real-valued functions on \(\mathbb{R}^3 \), then \[ f=h(g_1,g_2,g_3) \] is the function such that \[ f\boldsymbol{p}=h(g_1(\boldsymbol{p}),g_2(\boldsymbol{p}),g_3(\boldsymbol{p})) \] for all \( \boldsymbol{p}\). Express \( \partial f / \partial x \) in terms of \( x,y,z \) if \( h=x^2-yz \) and
4.a. \[ \frac{\partial }{\partial x }f= \frac{\partial }{\partial x } ( h(x+y,y^2,x+z) ) \] \[ \frac{\partial }{\partial x }f= \frac{\partial }{\partial x } ( x^2-yz ) \] We take here \( x+y \) as the \(x\) argument for \(h\), \(y^2\) for the \(y\) component, etc. \[ \frac{\partial }{\partial x }f= \frac{\partial }{\partial x } ( (x+y)^2-(y^2)(x+z) ) \] \[ = \frac{\partial }{\partial x } ( (x^2+y^2+2xy)-(y^2x+zy^2) ) \] \[ = \frac{\partial }{\partial x } ( (x^2+y^2+2xy)-y^2x-zy^2 ) \] \[ = 2x+2y-y^2 \]
4.b. \[ \frac{\partial }{\partial x }f= \frac{\partial }{\partial x } ( h(e^z,e^{x+y},e^x) ) \] \[ \frac{\partial }{\partial x }f= \frac{\partial }{\partial x } ( x^2-yz ) \] We make the similar replacements for the \(x\) argument for \(h\), etc. \[ \frac{\partial }{\partial x }f= \frac{\partial }{\partial x } ( (e^z)^2-(e^{x+y})(e^x) ) \] \[ = \frac{\partial }{\partial x } ( e^2z-e^{2x+y} ) \] \[ = -2e^{2x+y} \]
4.c. \[ \frac{\partial }{\partial x }f= \frac{\partial }{\partial x } ( h(x,-x,x) ) \] \[ \frac{\partial }{\partial x }f= \frac{\partial }{\partial x } ( x^2-yz ) \] We make the similar replacements for the \(x\) argument for \(h\), etc. \[ \frac{\partial }{\partial x }f= \frac{\partial }{\partial x } ( (x)^2-(-x)(x) ) \] \[ = \frac{\partial }{\partial x } ( x^2+x^2) \] \[ = 2x^2 \]
In this section, we use the points to create vectors. A vector has two points to it. Some \(\boldsymbol{p}\) (point of application) where the tail starts and a \(\boldsymbol{v}\) (vector part) where the head ends. We will then label a vector as \( \boldsymbol{v_p} \) which is a vector that starts at \(\boldsymbol{p}\) and ends at \(\boldsymbol{v}\). The vector part is \(\boldsymbol{v}\) and the point of application is \(\boldsymbol{p}\).
Two vectors are equal \( \boldsymbol{v_p} = \boldsymbol{w_q}\) only if they have the same point of application. That is \( \boldsymbol{v} = \boldsymbol{w}\) and \( \boldsymbol{p} = \boldsymbol{q}\).
Tangent vectors with the same vector part but different application part are parallel. They point the same where but are located at different points in space.
Definition If the application part is the same for all vectors, these tangent vectors belong to the same tangent space. This means there is a different tangent space at every point in \( \mathbb{R}^3 \). A tangent space, \(T_p ( \mathbb{R}^3 ) \), is a vector space where all tangent vectors with the same point \( \boldsymbol{p} \) are located.
The tangent space, \(T_p ( \mathbb{R}^3 ) \), is itself considered a vector space. The same operations of \( \boldsymbol{v_p} + \boldsymbol{w_p} = (\boldsymbol{v}+ \boldsymbol{w})_p \) and \( c\boldsymbol{v_p} = (c\boldsymbol{v})_p \) are identical to the space of \( \mathbb{R}^3 \). They are isomorphic.
A vector field is a space in \( \mathbb{R}^3 \) isomorphic to \(T_p ( \mathbb{R}^3 ) \) that has a vector arrow at each location in it. An example from physics is an electric field or a gravitational field.
Vector fields can be added just like functions. Adding can be thought of as vectorially adding vectors at each point in space all at the same time.
That is, as long as the same point \(\boldsymbol{p}\) is the argument, we can add \[ V(\boldsymbol{p})+W(\boldsymbol{p})= (V+W)(\boldsymbol{p}) \]
Multiplication by a scalar function works the same way as multiplying by a scalar with a vector. \[ (fV(\boldsymbol{p})) = f(\boldsymbol{p})V(\boldsymbol{p}) \] for all \(\boldsymbol{p}\).
Definition In a tangent space, we can define a natural frame field so that this tangent space is a vector space spanned by special, unit length vector fields on \( \mathbb{R}^3 \) which are \[ U_1(\boldsymbol{p}) =(1,0,0)_p \] \[ U_2(\boldsymbol{p}) =(0,1,0)_p \] \[ U_3(\boldsymbol{p}) =(0,0,1)_p \] They are defined like this for every point \(\boldsymbol{p}\) in \( \mathbb{R}^3 \). The unit field vector \(U_i (i=1,2,3) \) is in the positive \(x_i\) direction.
Lemma If \(V\) is a vector field on \( \mathbb{R}^3 \), then three real-valued functions, \(v_1,v_2,v_3 \) on \( \mathbb{R}^3 \) determine the vector field \[ V=v_1 U_1+v_2 U_2+v_3 U_3 \] \(v_1,v_2,v_3 \) are the Euclidian coordinate functions of V.
Proof The vector field \(V\) has a tangent vector \(V(\boldsymbol{p}) \) at \(\boldsymbol{p}\). because the vector part of the field depends on \(\boldsymbol{p}\), so that \(v_1(\boldsymbol{p}),v_2(\boldsymbol{p}),v_3(\boldsymbol{p}) \). This means the Euclidian coordinate functions are real-valued functions on \( \mathbb{R}^3 \). So, \[ V(\boldsymbol{p}) = (v_1(\boldsymbol{p}),v_2(\boldsymbol{p}),v_3(\boldsymbol{p})) \] \[ V(\boldsymbol{p}) = v_1 (\boldsymbol{p})(1,0,0)_p +v_2(\boldsymbol{p}) (0,1,0)_p+v_3(\boldsymbol{p})(0,0,1)_p \] \[ V(\boldsymbol{p}) = v_1 (\boldsymbol{p})U_1 +v_2(\boldsymbol{p}) U_2+v_3(\boldsymbol{p})U_3 \] We can rewrite this compactly as \[ V = \sum v_i U_i \] where \(i=1,2,3\). \( \blacksquare \)
1. Let \(\boldsymbol{v}=(-2,1,-1)\) and \(\boldsymbol{w}=(0,1,3) \).
1.a. We rewrite each vector first as linear combinations of the natural frame field. Then we substitute them in the formula and compute the resulting tangent vector. \[ \boldsymbol{v_p} = (-2,1,-1) = -2 (1,0,0)_p +1(0,1,0)_p - 1 (0,0,1)_p \] \[ \boldsymbol{w_p} = (0,1,3) = 0 (1,0,0)_p +1(0,1,0)_p + 3 (0,0,1)_p \] Since we can add point-wise with any point, \[ 3\boldsymbol{v_p}-2\boldsymbol{w_p} = (3\boldsymbol{v}-2\boldsymbol{w})_p \] So that, \[ 3\boldsymbol{v_p} = 3(-2,1,-1) = -6 (1,0,0)_p +3(0,1,0)_p - 3 (0,0,1)_p \] \[ -2\boldsymbol{w_p} = -2(0,1,3) = -2(0,1,0)_p -6 (0,0,1)_p \] \[ 3\boldsymbol{v_p}-2\boldsymbol{w_p} = -6 (1,0,0)_p +3(0,1,0)_p - 3 (0,0,1)_p -2(0,1,0)_p -6 (0,0,1)_p \] Collect like terms, i.e. add together the components, \[ 3\boldsymbol{v_p}-2\boldsymbol{w_p} = -6 (1,0,0)_p +1(0,1,0)_p - 9 (0,0,1)_p \] \[ 3\boldsymbol{v_p}-2\boldsymbol{w_p} = -6 U_1(\boldsymbol{p}) +1U_2(\boldsymbol{p}) - 9 U_3(\boldsymbol{p}) \]
1.b. Need to use drawing program here.
2. Let