Of Existence and Brains

Cogito, ergo sum!

Many of us are familiar with René Descartes' statement 'Cogito ergo sum' which is translated to 'I think, therefore I am'. Many of us take this as 'proof' of our existence, but it's not nearly that simple. First, we'll take a look at how he arrived at this.

Descartes arrived at this by systematically doubting everything that could be doubted and he could doubt no more. He realised that what he could not doubt was his own existence because to doubt his own existence, he would necessarily have to exist. Since he had set out with the building philosophy from the ground up, he took this as an axiom and said 'Je pense donce je suis'(Which is quoted verbatim from the Discourse on the Method), which is what I had quoted earlier in Latin and English.

Unfortunately, this statement is not logical inference of the existence of the self. Some of you must have already guessed the basic problem with this statement: The word 'I'.

Take the sentence, 'I think therefore I am'; I exist because I can think(and therefore doubt my own existence). But we are already presupposing the existence of 'I', making a statement about 'I' and then reaffirming that 'I' exists and that is a triviality. To think at all, one must exist in the first place!(Following Kierkegaard, Nietzsche and Russell)

Master of logical consistency

Now, one can actually go further and deny the existence of the self, but is it possible to deny thoughts? I'm not sure it is because even if you are a part of a larger consciousness, your thoughts being just images of that of something bigger and beyond comprehension, the thoughts still exist in some form. If we weaken the statement to 'Thoughts exist', the objection doesn't stand.

However, it is undeniable that there seems to be some truth to what Descartes says and his argument has definite intuitive appeal; it does feel right. So how do we go about justifying it?

The first step would be to stop treating 'I am' as a logical consequence of 'I think', but as a product of a process. The irrefutability of my own existence is a product of the process of my questioning it.

How? Let me say, 'I do not exist'. There's nothing logically wrong with the sentence, but there is something wrong with me stating it. For me to state this sentence, I would necessarily have to exist. If I try convincing you of the fact that I do not exist, by doing so, I am contradicting myself. This is an example of what Jaakko Hintikka calls an existentially inconsistent sentence. When you say 'I do not exist', you are not treating 'I' as an arbitrary quantity, but you are referring to yourself. When you make any self-referential statement, you necessarily acknowledge your own existence.

Who said old Finnish Logicians don't look cool? Jaakko Hintikka disproves!

But say all that exists are your thoughts and everything else is an illusion, what then? Isn't that a dead end too? Probably the simplest way to refute this is to rely on a principle that was championed by Bertrand Russell called Occam's Razor. This basically states that if there are two competing hypotheses, the one making less suppositions is the better hypotheses. It turns out that accepting reality as we perceive it requires far less suppositions than if we were to interpret these as illusions(Refer to Bertrand Russell's Problems of Philosophy).

But there's another approach to this problem and I'll explain it next:

"Your mind makes it real"

Let's modernise the setting, let's assume the I'm a brain in a vat which is being fed all sorts of information that corresponds to my senses by a computer. It definitely helps if you've seen The Matrix.(We're following Hilary Putnam's arguments).

Yes you are.

But first, let's consider a few things; If say, an alien from a different planetary system comes to Earth, sees our flora and draws a tree; without much objection, we can say that the picture does indeed refer to a tree. Now, say the alien has never been to Earth and draws a rather accurate picture of a tree quite randomly; can one say that the picture refers to an object or concept that it has never experienced before? According to Putnam, the answer is no.

This is what is termed as causal constraint and the principle is illustrated as follows:
'A term refers to an object only if there is an appropriate causal connection between that term and the object'

Now suppose a brain-a-vat(BIV for short) talks of 'trees'(note that the BIV cannot see a tree in your sense, and even when it talks of trees, it seems to speak because it's a brain in a vat -_- ), by causal constraint, it cannot be referring to a tree as we know it(and trees may not even exist), so what does it mean by the tree token? Putnam says that this can be one of three things:

1. 'Trees in the image' i.e. the experiences of the brain.
2. Neural impulses that stimulate the brain causing it to have experiences similar to those of a normal human while it sees trees.
3. The computer programme that is responsible for the stimuli and experiences.

   

   Trapped in a vat? Who you gonna call? Hilary Putnam!

Before Putnam, if the BIV had stated 'Here is a tree', it would have been taken that the BIV's token statement was false since we would have assumed that the BIV's token of a tree was referring to real trees. However, if we are to assign truth values to the BIV's statement on the basis of Putnam's assignments of possibilities, it would come out to be true. This is because its sentences express beliefs that are true internally in the brain's world. If you know no external reality, then what you experience is true inside your 'illusion'. 
Now, we are prepared to examine Putnam's arguments. Let us denote distinguish statements of the BIV from that of ours by suffixing the BIV's language with a '*'. 

1. Either I am a BIV(speaking vat-language) or I am not a BIV(speaking human language)
2. If I am a BIV(speaking vat-language),  then my utterances of ‘I am a BIV’ are true iff I am a brain* in a vat*.
3. If I am a BIV (speaking vat-English), then I am not a brain* in a vat*.
4. If I am a BIV (speaking vat-English), then my utterances of ‘I am a BIV’ are false. [inferred from 2. and 3.]
5. If I am a non-BIV (speaking English), then my utterances of ‘I am a BIV’ are true iff I am a BIV.
6. If I am a non-BIV (speaking English), then my utterances of ‘I am a BIV’ are false. [trivially, from 5.]
7. My utterances of ‘I am a BIV’ are false. [from 1., 4. and 6.]
8. My utterances of 'I am not a BIV' are true. [contrapositive of 7.]
9. My utterances of 'I am not a BIV' are true iff I am not a BIV
10. Hence, I am not a BIV

And as promised, I deduced that under certain assumptions, it is impossible to be a BIV. 

Contrary to what it might seem at first sight, this is actually one of the strongest critiques of metaphysical realism. According to the Stanford Encyclopaedia of Philosophy:

To metaphysical realism, the world is as it is independently of how humans take it to be. The objects the world contains, together with their properties and the relations they enter into, fix the world's nature and these objects exist independently of our ability to discover they do. Unless this is so, metaphysical realists argue, none of our beliefs about our world could be objectively true since true beliefs tell us how things are and beliefs are objective when true or false independently of what anyone might think.

If one were to subscribe to the position of metaphysical realism, he would be forced to admit the plausibility of a 'grand illusion' like the Brain-in-the-Vat thought experiment. However, Putnam showed that such a position is nonsensical even if we take the premises to be true.

Of course, there exist multiple counter-arguments and defenses of metaphysical realism. What is interesting to note is that Hilary Putnam himself was once a proponent of metaphysical realism. When he had published his critiques of metaphysical realism(the BIV experiment and the Model-Theoretic argument, which relies heavily on mathematical logic), he had changed his position to what he calls, Internal Realism. This internal realism resembles the metaphysics of Immanuel Kant on quite a few counts and has been termed as 'idealism in the guise of realism' by many. But it is beyond the scope of this essay to examine any of these.

References:

1. Key Philosophical Writings- René Descartes- Wordsworth Editions
2. Problems of Philosophy- Bertrand Russell- Oxford University Press
3. Cogito, ergo sum: Inference or Performance?- Jaakko Hintikka- Philosophical Review 71-1962
4. Reason, History, Truth- Hilary Putnam- Cambridge University Press
5. Stanford Encyclopedia of Philosophy- plato.stanford.edu

Of Hairy Balls, Fixed Points and Coffee

The title of this post must be inducing a giggle or two from the readers, and yes, you read that right; I will be discussing the 'Hairy Ball Theorem' in this post.

So what is the Hairy Ball Theorem? Formally, it states that an even-dimensional sphere does not possess any continuously differentiable field of unit tangent vectors.

In simpler terms, it basically states that you cannot comb a hairy ball flat without creating a cowlick(a point at which the derivative does not exist, a.k.a. a singularity). No! Not the kind of hairy ball you were thinking of, more like this:

.....or this:

.......and in case you were wondering, this is a cowlick:

Hang on! The coconut is not a ball!

No need to fret, this is topology and we can deform a sphere into a coconut-shape without making cuts or gluing stuff to it. Formally, we say that the coconut-shape is homeomorphic to a 2 dimensional sphere.

Wait a sec! Weren't we talking about 3D spheres just now?

Nope, what we normally call a sphere is mathematically called a 2-sphere for matters of convenience. The 1-sphere is the circle and the 0-sphere is a point. A 3-sphere would basically be a 4 dimensional object.

So coming back to our discussion, if you think about it for a while, you can convince yourself that it is impossible to comb a hairy ball flat.

But why just talk of even dimensional spheres? That's because this fails for the odd-dimensional case, if you need to convince yourself of this, draw a circle and start drawing tangents in an anti-clockwise(or clockwise) direction. You should be able to see that if it were a 'hairy circle', one might easily comb these hairs down in a clockwise or anti-clockwise direction without creating a cowlick. 

However, in mathematics, conviction isn't enough; there might very well be a way to comb a hairy ball flat which we haven't been able to imagine just yet. And moreover, although we have established that a hairy circle can be combed flat, we still haven't established this is true for higher odd dimensional spheres.

But before I go into a formal proof of the theorem, let me talk about its significance and history.

The hairy-ball theorem is actually the consequence of a very important in theorem in topology, first discovered by Henri Poincaré, called the Poincaré-Hopf theorem. This theorem relates the 'index' of a vector field(an analytic concept) to the Euler characteristic of a closed surface(a topological invariant), thereby bringing in a relation in between 2 fields of mathematics which were thought to be unrelated.

Henri Poincaré

However, the Hairy Ball Theorem was proved in generality by the Dutch Mathematician, and father of the Intuitionist School of Mathematics, Luitzen Egbertus Jan Brouwer.

It came about as he was trying to find a proof of what is now known as Brouwer's Fixed Point theorem. The story goes that as he was stirring sugar into a cup of coffee. As he was stirring, he noticed that there always seemed to be a point on the surface which wasn't moving. He concluded that at any given instant, there was always a point which wasn't in motion, even though this point could change in different instances.

What Brouwer's Fixed Point Theorem states is that every continuous mapping from an n-dimensional disk to itself contains a fixed point i.e. a point that remains invariant under the map. 

This theorem can be deduced from the Hairy Ball Theorem, and I will come to this in my final section.

L.E.J. Brouwer, proving Alfréd Rényi's statement, ''A mathematician is a machine for turning coffee into theorems''

The importance of the fixed point theorem(and its analogues) cannot be underestimated.... It has applications everywhere from mathematics, physics and logic; to economics and biology! Indeed, the Brouwer fixed point theorem was central to the proof of existence of general equilibrium in market economies, for which  Kenneth Arrow and Gérard Debreu received the Nobel Prize in Economics in 1983. It was also used in the original proof of the Nash Equilibria, for which John Forbes Nash Jr. also won the Nobel Prize in Economics.

Kenneth Arrow

G. Debreu 

An interesting physical application of the Hairy Ball theorem is that of wind patterns on planets. Now, since the atmosphere is always in motion due to the planet's rotation, this creates a vector field. By the Hairy Ball Theorem, there must be a position where there is no motion of the atmosphere and this corresponds to the eye of a cyclone or an anti-cyclone. Of course, this effect isn't always observable due to the fact that our atmosphere has layers, but even so, in every layer, there must be a point where no atmospheric motion takes place.

Jupiter with its massive Great Red Spot, a massive anti-cyclonic system

Brouwer's original proof relied on methods from algebraic topology, and that would be beyond the scope of this article. 

Luckily for us, a rather elementary proof of the Hairy Ball Theorem was discovered by John Willard Milnor(the 2011 Abel Prize winner, also a Fields Medalist and Wolf Prize Winner). This proof relies on little else but some basic calculus and analysis.

John Milnor, probably the most important topologist in the latter half of the 20th century

The proof will closely follow that of Milnor, we will first prove the Hairy Ball Theorem and then, using another statement of it, which will be used to prove Brouwer's Fixed Point Theorem.

And now for the proof:

A sphere \(S^{n-1}\) is the set of all vectors in

\(\mathbb{R}^n\) ,\(\alpha=(a_1, a_2,.....a_n)\) such that \(\|\alpha\|=1\)

A vector \(\beta(\alpha)\) in \(\mathbb{R}^n\) is tangent to \(S^{n-1}\) at \(\alpha\) if the inner product(i.e. the 'dot' product), \(\alpha\cdot\beta(\alpha)=0\)

If \(n-1\) is odd, the dot product in between \(\alpha\) and \(\beta(\alpha)\) will be null if \(\beta(\alpha)\) is chosen such that \(\beta(\alpha)=(a_2, -a_1,......, a_n, -a_{n-1})\), thus defining a differentiable vector field of tangent vectors on \(S^{n-1}\).

Now, we will break up the proof into two lemmas:

\(A\) is a compact region in \(\mathbb{R}^n\) and \(x:\to x+t\beta(\alpha)\) is a continuously differentiable vector field defined throughout the neighbourhood of \(A\), \(t\) is any real number.

Our 1st Lemma will be to prove that if \(t\) is sufficiently small, our function is a one-to-one function of \(A\) onto its image and its volume can be expressed as a polynomial function of \(t\)

Proof: \(\beta\) is continuously differentiable over the compact domain, by the mean value theorem, there exists a constant, \(c\) such that \(|\beta(x)-\beta(y)|\leq c|x-y|\)

We choose, \(|t|\leq c^{-1}\), then \(f\) is one-to-one because if \(f(x)=f(y)\), then \(x-y=t(\beta(x)-\beta(y))\) and the inequality \(\|x-y\|\leq |t|c\|x-y\|\) implies \(x=y\)

We write the matrix of the first derivatives of \(f\) as \(I+\frac{\partial\beta_i}{\partial x_j}\) where \(I\) is the identity matrix.

The determinant is a polynomial of \(t\) of the form \(1+tm_1(x)+....+t^nm_n(x)\), the coefficients are continuous functions of \(x\). For a sufficiently small \(|t|\), the determinant is strictly positive. 
The volume of the image region can be calculated by integrating over \(A\) and we see that it is a polynomial function of \(t\)

Now, if the sphere has a continuously differentiable field \(\alpha:\to \alpha+t\beta(\alpha)\) of unit tangent vectors, for any real number \(t\) this function has a length given by \(\sqrt{1+t^2}\)

Our 2nd Lemma will need to prove that if \(t\) is sufficiently small, the function maps the unit sphere in  \(\mathbb{R}^n\) onto the sphere of radius  \(\sqrt{1+t^2}\)
Proof: We assume that \(n>2\) since we've already settled the case for \(S^1\). If \(t\) is sufficiently small, the matrix of 1st derivatives of \(f\) is non-singular throughout the compact region \(A\). By the Inverse Function Theorem, \(f\) maps open sets in the interior of \(A\) to open sets. Hence, the image of the unit sphere is a relatively open subset of the sphere of radius  \(\sqrt{1+t^2}\). But since the image is compact, it is closed and since the image is both closed and open, it must be the entire sphere of radius  \(\sqrt{1+t^2}\).

With these two lemmas in hand, we can now prove the Hairy Ball Theorem.

As region \(A\), we take the space in between the two concentric spheres which is defined by the inequalities \(p\leq \|x\|\leq q\). We extend the vector field \(\beta\) throughout this region by setting \(\beta(r\alpha)=r\beta(\alpha)\) for \(p\leq r\leq q\), it follows that the mapping \(f(x)= x+t\beta(x)\) is defined throughout the region and maps the sphere of radius \(r\) onto the sphere of radius  \(r\sqrt{1+t^2}\) , given \(t\) is sufficiently small. Hence, it maps \(A\) onto the region in between the spheres of radius  \(p\sqrt{1+t^2}\) and  \(q\sqrt{1+t^2}\).

Now, we see that \(vol\ f(A)\)\(=\) \((\sqrt{1+t^2})^n\)\(vol\)\((A)\)

Now whenever \(n\) is odd, this expression is not a polynomial, which is a contradiction with Lemma 1.

Hence, we've proved the Hairy Ball Theorem.

Now another form of this theorem which follows as a corollary is that an even dimensional sphere does not admit any continuous field of non-zero tangent vectors.

I will not go through the pains of proving this, as TeXing is hard work. :P
But I will prove Brouwer's fixed point theorem using this corollary:
Brouwer's Fixed Point Theorem: Every continuous mapping \(f\) from a disk \(D^n\) to itself possesses at least one fixed point.

Proof: We prove by contradiction. If \(f(x)\neq x\) for all \(x\) in \(D^n\), then the formula \(\beta(x)=x-f(x)\) defines a non-zero vector field \(\beta\) on \(D^n\) which points outward everywhere on the boundary, i.e. \(\alpha\cdot\beta(\alpha)>0\) for every \(\alpha\) in \(S^{n-1}\)

If we set \(\gamma(x)=x-y(1-x\cdot x)/(1-x\cdot y)\), where \(y=f(x)\neq x\)

\(\gamma\) defines a non-zero vector field on \(D^n\) which points directly outward on the boundary i.e. \(\gamma(\alpha)=\alpha\) for all \(\alpha\) in \(S^{n-1}\)

\(\gamma(x)=x\) when \(x\cdot x=1\) and since the denominator never vanishes, the expression depends continuously on \(x\). When \(x\) and \(y\) are linearly independent, \(\gamma(x)\neq 0\) and when they are linearly dependent, \((x\cdot x)y\)=\((x\cdot y\)x\) implies \(\gamma(x)=(x-y)/(1-x\cdot y)\neq 0\)

Now, we will appeal to our geometric intuition for the sake of simplicity:
We take the our new vector field \(\gamma(\alpha)\) to the Southern Hemisphere of our unit sphere \(S^n\) in \(\mathbb{R}^{n+1}\). We identify \(\mathbb{R}^n\) with the hyperplane passing through the equator, \(x_{n+1}=0\) , we use stereographic projection from the North Pole to map every point \(x\) of \(D^n\), to a point \(s(x)=\alpha\) of the southern hemisphere \(a_{n+1}<0 .="" br="" nbsp="">

Applying the derivative of the mapping s at x to \(\gamma(x)\), we get a corresponding tangent vector \(\theta(\alpha)\) to \(S^n\) at the image point \(s(x)=u\) . We obtain a non-zero tangent vector field \(\theta\) in this manner, and not surprisingly, it is oriented towards the North Pole. This is because at every point of the equator \(\alpha=s(\alpha)\), \(\gamma(\alpha)=\alpha\) points outwards.


Again using stereographic projection, from the South Pole, the vector field \(-\gamma(x)\) corresponds to a vector field on the northern hemisphere that is also directed towards the North Pole. 
Now, if we join these two vector fields together, we see that the resultant vector field is directed towards the North Pole. If \(n\) is even, by the Hairy-Ball Theorem, this is impossible and hence we have a contradiction.

The visualisation of a stereographic projection; adapted from Milnor's original

Hence, our assumption that \(f(x)\neq x\) for all \(x\) in \(D^n\) is false. Or in other words, there must exist a fixed point.

We've proved it for the cases where \(n\) is, but even in the case that \(n\) is odd, this follows from our proof. In this case, \(D^n\subset D^{n+1}\). Hence any map from \(D^n\) to itself is also a map from \(D^{n+1}\) to itself and if such a map does not have a fixed point, it violates the fixed point theorem that we have established, as \(n+1\) is even. And hence this theorem applies to odd \(n\) as well.

References:

The proof of the Hairy-Ball Theorem and Brouwer's Fixed Point Theorem has been adapted from the following paper:

• [JM1] Milnor, John-Analytic proofs of "hairy ball theorem" and Brouwer fixed-point theorem-The American Mathematical Monthly- Vol. 85, No. 7 (Aug. - Sep., 1978), pp. 521-524

For a discussion of the Poincaré-Hopf theorem and an introduction to differential topology, see:

• [JM2] Milnor, John- Topology from the Differentiable Viewpoint- Princeton University Press

For the definitions and theorems from analysis, such as the meanings of open set, closed set, compactness and the inverse function theorem:

• [WR] Rudin, Walter- Principles of Mathematical Analysis- McGraw-Hill
• [TA] Apostol, Tom- Mathematical Analysis- Narosa

For general topology and some algebraic topology:

• [MAA]- Armstrong, M.A.- Basic Topology- Springer-Verlag 
• [KJ] - Jänich, Klaus- Topology- Springer-Verlag
• [JRM]- Munkres, James- Topology- Prentice Hall

LaTex Test

\[ x = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a} \]\( x = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a} \) \( x = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a} \)