The Moral Arguments: A Rejoinder

Recently I began a delightful discussion/debate concerning two versions of the moral argument for the existence of God with my friend Tyler Dalton McNabb. His opening post can be read here. My first response can be read here and his followup response here.

In response to Tyler’s latest post, I’d like to start by pointing out that there are two central issues at play in this discussion. One regards the nature of morality itself and the other regards the “proper function” of our faculties and our ability to access truth.

As of right now, I hold firm to the position that the true nature of morality is undecidable. I contend that this is enough to neutralize the classical moral argument, so I’ll focus more on the epistemological version.

Now, while I hesitate to agree that belief in objective moral values and duties is warranted (since “warranted” is a technical term that still isn’t adequately understood), I will admit that belief in objective moral values and duties can be reasonable. Where I think the epistemological argument truly fails is premise (1):

If NE is true, belief in objective moral values and duties cannot be warranted.

Tyler conceded that premise (1), so stated, is false, since it is possible on NE that our faculties be such that they can access truth and form corresponding true beliefs. What Tyler seems to resort to is something like the following:

Let \alpha represent the actual world (at least with respect to our perspective). Furthermore, let W be the event that belief in objective moral values and duties is warranted on NE in \alpha. Then

P(W) < \frac{1}{2}

which says that the probability of W being the case is less than 50%. If this is the case, then it is more reasonable to think that one is not warranted in believing in objective moral values and duties in \alpha. But is this really the case?

First, I’m not sure that there is a well-defined probability measure here. For instance, if there are infinitely many universes, then infinitely many of them have beings with the right cognitive equipment to access truth.

Second, Tyler says that he sees no reason to think that our cognitive faculties are reliable. This is strange to me, since I see every (well maybe not every) reason to think that our faculties are reliable. Why?

(a) This may be a necessary part of consciousness. That is, being conscious may very well entail the ability to form true beliefs and to check them on some level.

(b) We can test our faculties against real life. The fact that we don’t die and the fact that we can successfully navigate our world environment attest to the fact that our brains produce true beliefs.

For these reasons, I think the epistemological argument fails to go through. It is entirely possible, even likely, that our cognitive faculties are reliable even on NE. At this point, I’ll now pass the ball back to Tyler.

Bill Nye vs Ken Ham: A Break Down of the Debate – Part 1

Lately the interweb has been abuzz with anticipation and now reflection upon the debate between Bill Nye and creationist Ken Ham of Answers in Genesis. As expected, the debate was more polarizing than productive. As far as the actual debate, however, Nye won by a landslide. But that’s easy to say (or type), so let me tell you why Nye was the clear winner.

I’ll analyze each interlocutor in turn.

Ken Ham: Opening Statement

Despite my distaste for Ham, I’ll admit that he started out in a promising way.  He clearly laid out the debate topic, which was

Is creation a viable model of origins in today’s modern scientific era?

From here one would hope, if not expect, that one arguing in the affirmative would proceed to marshal evidence and justification for creation being a viable model of origins. Unfortunately, as soon as Ham sets up an appropriate trajectory, he deviates from it to address a completely different matter, namely, whether creationists can be scientists. He laments that secularists have hijacked the term “science” and then proceeds to introduce us to several scientists who happen to be Biblical creationists. While I can understand that Ham is trying to dispel a misconception, it seems that he is bordering on an argument from authority. The salient issue, however, is not whether creationists can be scientists, but whether the idea of creation itself is (a) scientific in any relevant sense and (b) whether it holds up under scrutiny (of any kind).

Ham seems to get back on track (briefly) by rightly pointing out the need to define terms correctly. This is always important for fruitful communication. Beginning with “science”, Ham distinguishes between observational science and what he calls historical science. Observational science is what produces technology, whereas historical science obviously deals with past events that cannot be directly observed.  Under such a distinction, origins would clearly belong to the latter category. But is there really such a fine distinction? It is true that we cannot directly observe past events, but that does not mean that observational science is not involved in studying the past. What we observe are the clues of the present. Past events get encoded (more or less) into the future. We observe this and then attempt to work the “encoding path” backwards to the event in question.  This is how crime scene investigation works. Experts in arson can tell a lot about how a fire started, how fast it burned, where it started, and many other things by examining the remains of burnt houses and buildings. The event is in the past, but many important bits of information are encoded that can be deciphered.  Yes, Ham is right, no one was there to witness our origins, which is why we are trying to determine the best explanation and model.

Ham seems to think that this gives him an out. No one was there to witness these things, therefore it is legitimate and viable to simply base one’s views on (a very particular interpretation of) the Bible. While I will agree to the importance of world views and starting assumptions, Ham has not demonstrated the legitimacy of using an ancient religious text as a foundational starting point.

After rambling on about there being a difference in philosophical world views, Ham ends with a doozy.  He literally concludes that

Creation is the only viable model of historical science confirmed by observational science in today’s modern scientific era (emphasis mine).

Ummmm…. what?  At no point in his presentation did he present anything remotely close to evidence for this audacious conclusion.  He spent the entire time complaining about secularist hijackers and concluded that creation is the only viable model of origins.

So, Ham had a promising start, but went down in a ball of flames. In the next post, I’ll critique Nye’s opening presentation.

Graded Algebras and Classical Mechanics

Certain algebras possess an important classifying property known as a grading.  This essentially amounts to being able to express these algebras as a special kind of direct sum that respects multiplication.

Definition[Graded Algebra]:  An algebra A over a field \kappa is said to be graded if, as a vector space over \kappa, A can be written as the direct sum of a family of subspaces (A_{i})_{i\in \mathbb{N}} – i.e.

\displaystyle A = \bigoplus_{i\in \mathbb{N}}A_{i}

and such that multiplication behaves according to

A_{i}\cdot A_{j} \subseteq A_{i+j}, for all i,j\in \mathbb{N}

In classical mechanics, one deals with smooth manifolds. To understand what these are we need some definitions.

Definition[Chart]:  Let M be a manifold of dimension n.  Then a chart is a pair (U, \psi), where U\subset M is open and \psi: U\to V\subset \mathbb{R}^{n} is a homeomorphism to some open V.

The collection of all charts such that each x\in M belongs to some chart is called an atlas.  An atlas characterizes a manifold.  More formally,

Definition[Atlas]:  An atlas on a topological space T is a collection of charts \{(U_{\alpha},\varphi_{\alpha})\}_{\alpha\in I} where the U_{\alpha} are open sets that cover T, and for each index \alpha

\varphi_{\alpha}: U_{\alpha}\to \mathbb{R}^n

is a homeomorphism of U_{\alpha} onto an open subset of n-dimensional Euclidean space.

Definition[Smooth Atlas]: An atlas \mathcal{A} is called smooth if for all \varphi_{i}, \varphi_{j} we have that

\varphi_{j}\circ \varphi_{i}^{-1} : \varphi_{i}(U_{i}\cap U_{j})\to \varphi_{j}(U_{i}\cap U_{j})

is a diffeomorphism.


Definition[Diffeomorphism]: Given two manifolds M and N, a differentiable map f: M\to N is called a diffeomorphism if it is a bijection, and its inverse f^{-1}: N\to M is also differentiable.

Now, a manifold may not be generated by a unique atlas.  To account for this, we seek a preferred atlas known as a maximal atlas.  In defining such a thing, we need to know what it means for atlases to be compatible.

Definition[Compatibility]: Let \mathcal{A} and \mathcal{A}' be two smooth atlases.  Then \mathcal{A} and \mathcal{A}' are called compatible if and only if \mathcal{A}\cup \mathcal{A}' is again a smooth atlas. 

This notion of compatibility is actually an equivalence relation.  The union over an equivalence class of atlases gives us our maximal atlas.  If we denote an equivalence class by [\mathcal{A}], then we have

Definition[Maximal Atlas]:  A maximal atlas for a manifold M is the union of all smooth atlases in an equivalence class – i.e.

\mathcal{A}_{max} = \bigcup\{\mathcal{B} : \mathcal{B}\in [\mathcal{A}]\}

This maximal atlas is said to be the differentiable structure of the manifold.  We are now ready to understand the idea of a smooth manifold.

Definition[Smooth Manifold]:  A smooth manifold is a pair (M, \mathcal{A}_{max}) where M is a manifold and \mathcal{A}_{max} is a differentiable structure of M.

Smooth manifolds are important in classical mechanics because they allow us to do calculus on them.  Now, to return to the original topic, we are also interested in the graded algebra of smooth differentiable forms on a smooth manifold M with respect to the wedge product, which is denoted by

\displaystyle \mathcal{A}^{\bullet}(M) = \bigoplus_{k=0}^{n}\mathcal{A}^{k}(M)

– Differential Forms – 

For now, let’s restrict our focus to what are called differential 1-forms, or just 1-forms.  According to one source, “Informally, a differential form is what can be integrated along a path.”

[1-Forms]

Starting in \mathbb{R}^{2}, let U be some open subset of \mathbb{R}^{2} and let F, G be two real-valued functions defined on U.  Then an expression of the form

F(x,y)dx+G(x,y)dy

is called a differential 1-form on U.  A particularly important example of a differential 1-form is the total differential of a C^{1} real-valued function f defined on some open subset U.

Definition[Total Differential]:  Let f by a C^{1} real-valued function defined on an open subset U of \mathbb{R}^{2}.  Then its total differential df is defined by

\displaystyle df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy

This can also be expressed as a kind of dot product of vectors, namely \nabla f\cdot (dx,dy), where \nabla f is the gradient of f.  Note that in this case we have \displaystyle F(x,y) = \frac{\partial f}{\partial x} and \displaystyle G(x,y) = \frac{\partial f}{\partial y}.

1-forms can be defined similarly for higher dimensions.  For instance, a differential 1-form on an open subset U of \mathbb{R}^{3} is an expression of the form

F(x,y,z)dx+G(x,y,z)dy+H(x,y,z)dz

where F, G and H are real-valued functions on U.  Again, if f is a C^{1} function defined on U, then the total differential

\displaystyle df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz

is an example.

In general, a 1-form on an open subset of \mathbb{R}^n is an expression of the form

F_{1}(x_{1},x_{2},...,x_{n})dx_{1}+F_{2}(x_{1},x_{2},...,x_{n})dx_{2}+\ldots + F_{n}(x_{1},x_{2},...,x_{n})dx_{n}

or simply \displaystyle \sum_{i = 1}^{n}F_{i}(x_{1},x_{2},...,x_{n})dx_{i}.  If we let \boldsymbol{x} = (x_{1}, x_{2},..., x_{n}), then we can write a 1-form even more succinctly as

\displaystyle \varphi = \sum_{i=1}^{n}F_{i}(\boldsymbol{x})dx_{i}

So, if f is a C^{1} real-valued function defined on U, the the general total differential is

\displaystyle df = \sum_{i=1}^{n}\frac{\partial f}{\partial x_{i}}dx_{i}

Note that each dx_i is a 1-form.  These are defined by the property that for each vector \boldsymbol{v} = (v_{1},v_{2},...,v_{n})\in T_{x}\mathbb{R}^{n}

dx_{i}(\boldsymbol{v}) = v_{i}

Clearly, these form a basis for the 1-forms on \mathbb{R}^{n}.

Definition[Smooth 1-Form]:  A smooth 1-form \varphi on \mathbb{R}^{n} is a real-valued function on the set of all tangent vectors to \mathbb{R}^{n}, i.e. – 

\varphi : T\mathbb{R}^{n}\to \mathbb{R}

with the properties that

  1. \varphi is linear on the tangent space T_{x}\mathbb{R}^{n} for each x\in \mathbb{R}^{n} – i.e. if \boldsymbol{v}_{1}, \boldsymbol{v}_{2}\in T_{x}\mathbb{R}^{n} and c_{1},c_{2}\in \mathbb{R},then \varphi(c_{1}\boldsymbol{v}_{1}+c_{2}\boldsymbol{v}_{2}) = c_{1}\varphi(\boldsymbol{v}_{1})+c_{2}\varphi(\boldsymbol{v}_{2}).
  2. For any smooth vector field v = v(x), the function \varphi(v):\mathbb{R}^{n}\to \mathbb{R} is smooth.

A smooth 1-form on an n-dimensional manifold M is defined similarly.

Note:  Another source simply says that a smooth 1-form on an open subset U of \mathbb{R}^{n} is given by an expression

\displaystyle \varphi = \sum_{i=1}^{n}f_{i}dx_{i}

where f_{i}\in C^{\infty}(U).

To better understand the graded algebra of smooth differential forms on M with respect to the wedge product we need to understand that the 1-forms on \mathbb{R}^{n} are part of an algebra called the algebra of differential forms on \mathbb{R}^{n}.  Multiplication in this algebra is known as the wedge product and is denoted by the symbol “\wedge“.  It has the property of being skew-symmetric or anti-commutative:

dx_{i}\wedge dx_{j} = -dx_{j}\wedge dx_{i}

From this it follows that dx_{i}\wedge dx_{i} = -dx_{i}\wedge dx_{i}, which implies that dx_{i}\wedge dx_{i} = 0 (provided we are not working over a field with characteristic 2).  This means that the wedge product is alternating.

From here we can build up k-forms on \mathbb{R}^{n}.  If each summand of a differential form \varphi contains k dx_{i}‘s, then \varphi is called a k-form.

Note: Functions are considered to be 0-forms.

The set

\{dx_{i_{1}}\wedge \ldots \wedge dx_{i_{k}} : 1\leq i_{1} < i_{2} < \ldots < i_{k} \leq n\}

is a basis for the k-forms on \mathbb{R}^{n}.  Thus, every k-form can be expressed in the form

\displaystyle \varphi = \sum_{|I| = k}f_{I}dx_{i_{1}}\wedge \ldots \wedge dx_{i_{k}}

where I ranges over all multi-indecies I = (i_{1}, i_{2},...,i_{k}) of length k.

The Exterior Derivative

The exterior derivative is an operation that sends a k-form to a (k+1)-form.

Definition[Exterior Derivative (1-Form)]:  Let f be a 0-form (function).  Then its exterior derivative df is the 1-form

\displaystyle df = \sum_{i=1}^{n}\frac{\partial f}{\partial x_{i}}dx_{i}

Note that this corresponds to the total differential of f.

Definition[Exterior Derivative (k-form)]:  Let \varphi be a k-form.  Then its exterior derivative d\varphi is the (k+1)-form obtained from \varphi by applying $lated d$ to each of the functions involved in \varphi.

Definition[Directional Derivative]:  Let f be a differentiable function defined on a region R in \mathbb{R}^{n}.  Let \boldsymbol{v} be a vector based at the point p_{0}\in R.  Then the derivative of f along \boldsymbol{v}, denoted D_{\boldsymbol{v}}f, is defined as follows.  Let F(t) = f(p_{0}+t\boldsymbol{v}).  Then 

D_{\boldsymbol{v}}f = F'(0)

Theorem:  Let f be a differentiable function defined on a region R of \mathbb{R}^{n}.  Let \varphi be defined on tangent vectors to points of R by

\varphi(\boldsymbol{v}) = D_{\boldsymbol{v}}f

Then

\varphi = df

Proof:

Let \boldsymbol{v} be a vector based at a point p of R with coordinates (p_{1}, p_{2},..., p_{n}). Let \boldsymbol{v} have coordinates (v_{1}, v_{2}, ..., v_{n}).  Let \textbf{u}_{i} be the ith unit vector.  Then

\displaystyle df(\boldsymbol{v}) = \left(\sum_{i=1}^{n}\frac{\partial f}{\partial x_{i}}dx_{i}\right)\left(\sum_{i=1}^{n}v_{i}\textbf{u}_{i}\right)

=\displaystyle \sum_{i=1}^{n}\frac{\partial}{\partial x_{i}}f(p)v_{i}

Next, let’s evaluate \varphi(\boldsymbol{v}) = D_{\boldsymbol{v}}f = F'(0) where we recall that F(t) = f(p+t\boldsymbol{v}).  So, really, we should write \displaystyle F'(0) = \frac{\partial}{\partial t}f(0).  Now, F is actually a composition of two functions, F(t) = f(g(t)), where g(t) = p+t\boldsymbol{v} = (p_{1}+tv_{1}, p_{2}+tv_{2},...,p_{n}+tv_{n}).  If we now apply the chain rule, then we get

\displaystyle \frac{\partial}{\partial t}f(0) = \sum_{i=1}^{n}\frac{\partial}{\partial x_{i}}f(g(0))\frac{d}{dt}x_{i}(0)

But x_{i} = p_{i}+tv_{i}, so \displaystyle \frac{d}{dt}x_{i} = v_{i}.  Therefore,

\displaystyle \frac{\partial}{\partial t}f(0) = \sum_{i=1}^{n}\frac{\partial}{\partial x_{i}}f(p)v_{i}

 This finishes the proof.

The exterior derivative obeys both the Leibniz rule and the chain rule:

  • d(fg) = gdf+fdg         Leibniz Rule
  • d(h(f)) = h'(f)df        Chain Rule

If \phi is a p-form and \psi is  a q-form, then the Leibniz rule takes the form

d(\phi \wedge \psi) = d\phi \wedge \psi+(-1)^{p}\phi\wedge d\psi

Theorem:  For any differential form \phi,

d(d\phi) = 0

(Or more succinctly, d^2 = 0).

Proof:

First, let f be a function (i.e. 0-form).  Then

\displaystyle d(df) = d\left(\sum_{i=1}^{n}\frac{\partial f}{\partial x_{i}}dx_{i}\right)

\displaystyle = \sum_{i=1}^{n}d\left(\frac{\partial f}{\partial x_{i}}\right)dx_{i}

\displaystyle = \sum_{i=1}^{n}\left(\sum_{j=1}^{n}\frac{\partial f_{x_{i}}}{\partial x_{j}}dx_{j}\right)dx_{i}

[where \displaystyle f_{x_{i}} : = \frac{\partial f}{\partial x_{i}}]

\displaystyle =\sum_{i=1}^{n}\left(\sum_{j=1}^{n}\frac{\partial^{2} f}{\partial x_{i}\partial x_{j}}dx_{j}\right)dx_{i}

\displaystyle =\sum_{i,j}\frac{\partial^{2} f}{\partial x_{i}\partial x_{j}}dx_{j}\wedge dx_{i}

\displaystyle =\sum_{i<j}\left(\frac{\partial^{2}f}{\partial x_{j}\partial x_{i}} - \frac{\partial^{2}f}{\partial x_{i}\partial x_{j}}\right)dx_{i}\wedge dx_{j} = 0

since mixed partials commute.  Now, since dx_{i} actually means d(x_{i}), where x_{i} is the i-th coordinate function, we have that d(dx_{i}) = 0.  Let

\displaystyle \phi = \sum_{|I| = k}f_{I}dx_{i_{1}}\wedge \ldots \wedge dx_{i_{k}}

Then

\displaystyle d(d\phi) = \left(\sum_{|I|=k}df_{I}\wedge dx_{i_{1}}\wedge \ldots \wedge dx_{i_{k}}\right)

\displaystyle = \sum_{|I|=k}\Big(d(df_{I})\wedge dx_{i_{1}}\wedge \ldots \wedge dx_{i_{k}}-df_{I}\wedge d(dx_{i_{1}})\wedge \ldots \wedge dx_{i_{k}}+\ldots\Big)

[By the Leibniz Rule]

Per the reasoning above with functions, this is clearly 0.

Much of the above has been described in terms of differential forms on \mathbb{R}^{n}.  Most of it transfers fairly seamlessly to general manifolds (i.e. the concepts are defined similarly).  So, to tie this all back to the graded algebra given above, we should understand \mathcal{A}^{0}(M) to be C^{\infty}(M) (i.e. the 0-forms on M) and \mathcal{A}^{k}(M) to be the algebra of smooth differential k-forms on M. We’ll look more at graded algebras in the next post.