Product of an Antisymmetric Tensor and a Symmetric Tensor

A well-known homework problem from General Relativity and Tensor Analysis textbooks asks the student to prove that:

$A^{\alpha \beta }B_{\alpha \beta } = 0$,

if A is an antisymmetric $\left ( \frac{2}{0} \right )$ tensor and B is a symmetric $\left ( \frac{0}{2} \right )$ tensor.

Now, this problem will do your head in if you treat the above equation as a matrix product. Remember that we're looking at Einstein Summation Notation here and all will be fine. If we sum on repeated indices, we can work things out in the following way:

$A^{\alpha \beta }B_{\alpha \beta } =-A^{\beta\alpha}B_{\beta\alpha}$ (antisymmetry of A and symmetry of B)     (1)
$=-A^{\mu \nu }B_{\mu \nu }$ (relabeling dummy indices)    (2)
$=-A^{\alpha \beta }B_{\alpha \beta }$ (relabeling dummy indices again)    (3)

Now, if you're not comfortable at this point with just relabeling indices, it's instructive to work out the expressions on both the L.H.S. and the R.H.S. of the equation:
$A^{\alpha \beta }B_{\alpha \beta } = -A^{\beta\alpha}B_{\beta\alpha}$    (1)

L.H.S.: $A^{\alpha \beta }B_{\alpha \beta } = A^{11}B_{11}+A^{12}B_{12}+A^{21}B_{21}+A^{22}B_{22}+\cdot \cdot \cdot$
R.H.S.: $-A^{\beta\alpha}B_{\beta\alpha} = -A^{11}B_{11}-A^{21}B_{21}-A^{12}B_{12}-A^{22}B_{22}+\cdot \cdot \cdot$
Rearranging terms and factoring out a negative one on the R.H.S. gives us:
$-A^{\beta\alpha}B_{\beta\alpha} = -1\ast \left [  A^{11}B_{11}+A^{12}B_{12}+A^{21}B_{21}+A^{22}B_{22}+\cdot \cdot \cdot \right ]$
Which shows that:
$-A^{\beta\alpha}B_{\beta\alpha}=-A^{\alpha \beta }B_{\alpha \beta }$

So, now we go back to (3) above:
$A^{\alpha \beta }B_{\alpha \beta }=-A^{\alpha \beta }B_{\alpha \beta }$
$\Rightarrow 2A^{\alpha \beta }B_{\alpha \beta }=0$
$\Rightarrow A^{\alpha \beta }B_{\alpha \beta } =0$    Q.E.D.

To Survive, Perchance To Thrive

One's raison d'etre is exactly one's own. I'm reminded of a time when I was at a friend's home and his sister complained about not knowing her purpose in life. to which her father responded sharply, "Your purpose in life is to not die. Everything else is gravy."

Gravy...

To reduce it to the most basic level, human beings seek to survive and to thrive. Indeed, faced with an inability to thrive on his or her own terms, one might contemplate suicide. We know this from being observers of humanity. Of course, 'on one's own terms' is key here. I've been told on several occasions that my spartan lifestyle is quite unappealing. I, of course, find it satisfying, but I can accept that it wouldn't be for most.

We're binary thinkers by nature, yes or no, right or wrong, good or bad, one or zero... But any rational human being should accept that these are relative concepts, not absolute, universal ones. Still, we'd like to put it in some kind of mathematical framework if we could.

Understanding survival is simple; there are two states: life and death. You're either in one or the other as far as we can tell from our human viewpoint. For the sake of simplicity, let's assume that the following is true:

Signs of death or strong indications that a warm-blooded animal is no longer alive are:

• Cessation of breathing - a
• Cardiac arrest (no pulse) - b
• Pallor mortis, paleness which happens in the 15–120 minutes after death - c
• Livor mortis, a settling of the blood in the lower (dependent) portion of the body - d
• Algor mortis, the reduction in body temperature following death. This is generally a steady decline until matching ambient temperature - e
• Rigor mortis, the limbs of the corpse become stiff (Latin rigor) and difficult to move or manipulate - f
• Decomposition, the reduction into simpler forms of matter, accompanied by a strong, unpleasant odor - g

We can then say that if none of the above conditions are met for a certain person, then that person is alive, meaning he or she is surviving.

So now, again for the sake of simplicity, let's define a state of death where only if all of the above conditions are true, then there is a state of death. And, let's call this state 'non-survival.' Sounds silly, but it makes things simple. We can represent this as a variable x, where x is a state function, x(a, b, c, d, e, f, g). If any of the conditions a, b, c, d, e, f or g are true, we assign that condition variable a value of 1, and if not true, a value of 0. For example, if a person is still breathing, a = 0, and if they're not breathing, a = 1. So, if we have x(1, 1, 1, 1, 1, 1, 1), we have a state of death. We can define a general life/death state as

                                    $\delta\left ( x \right ) = \begin{Bmatrix}
                                    0, & x(1, 1, 1, 1, 1, 1, 1)\\
                                    1,& x\neq x(1, 1, 1, 1, 1, 1, 1)
                                    \end{Bmatrix}$

where $\delta \left ( x  \right )= 1$ means survival, and
$\delta \left ( x  \right )= 0$ means non-survival.

Can we construct a similar function for thriving? And, if we can do that, can we construct a single, combined function that expresses the two (survival and thriving) mathematically?

Bijections and Finite Sets: Proof by If And Only If

Here's a proof that requires a bit of subtlety...

Prove that a nonempty set $T_{1}$ is finite if and only if there is a bijection from  $T_{1}$ onto a finite set  $T_{2}$.

We can use the following:

• Definition     (a)  The empty set is said to have zero elements.
• (b)  If $n\in \mathbb{N}$,  a set S is said to have n elements if there exists a bijection from the set         $\mathbb{N}_{n}$ := {1, 2, ..., n} onto S.
• (c)  A set S is said to be finite if it is either empty or has n elements for some $n\in \mathbb{N}$.
• (d) A set S is said to be infinite if it is not finite.

Step 1: If
Assume there is a bijection from $T_{1}$ onto $T_{2}$. We're given that $T_{2}$ is finite. We know that by definition, every element of one set is paired with exactly one element of the other set, and every element of the other set is paired with exactly one element of the first set ($T_{1}$). Since $T_{2}$ is finite, $T_{1}$ is finite.

Step 2: Only If
Assume $T_{1}$ is finite set, which is nonempty as described in the original problem statement. This means that by the definition above, $T_{1}$ has n elements for some $n\in \mathbb{N}$ and there exists a bijection from the set $\mathbb{N}_{n}$ := {1, 2, ..., n} onto $T_{1}$. Let's call that bijection f.

We also are given that $T_{2}$ is a finite set. Assuming that $T_{2}$ is not empty, we know by the definition above that there exists a bijection from the set $\mathbb{N}_{n}$ := {1, 2, ..., n} onto $T_{2}$. Let's call that bijection g.

The composition $g\circ f$ would map $T_{1}$ into $T_{2}$, and since the composition of two bijections is a bijection, $g\circ f$ is a bijection from $T_{1}$ onto $T_{2}$.

QED 

Exponents and Factorials: Mathematical Induction Proof

Easy one here...

Prove that $2^{n}< n!$ for all $n\geq 4$, $n\in \mathbb{N}$.

Step 1: Show P(4) is true.
Our base case is k = 4. $2^{4}<4!$ $\Rightarrow 16 < 24$. OK.

Step 2: Show that (if we assume P(k) is true), P(k+1) is true. (Induction hypothesis). This means we should have $2^{k+1}<(k+1)!$.

$$2^{k+1}<(k+1)!$$
can be rewritten as
$$2\cdot2^{k}<(k+1) k!$$

For $k\geq 4$, 2 < k + 1, and by the induction hypothesis, we assume $2^{k}< k!$, so we can see that
$$2^{n}< n!$$
for all $n\geq 4$, $n\in \mathbb{N}$

QED

Composition Of Two Bijections Is a Bijection: Proof

A common exercise from an "Introduction To Real Analysis" course is to show that

1. if $f: A\rightarrow B$ is a bijection, and
2. if $g: B\rightarrow C$ is a bijection

then the composite $g\circ f$ is a bijective map of A onto C.

We just need to show that $g\circ f: A\rightarrow C$ is one to one and onto.

$$g\circ f = g(f(x))$$
$$f(x)=y     (y\in B)$$
$$g(y)=z    (z\in C)$$

First, show the composite is injective (one to one):
by 1, we know that $f(x_{1})=f(x_{2})$ means $x_{1}=x_{2}$ (f is one to one - injective).
by 2, we know that $g(f(x_{1}))=g(f(x_{2}))$ means $f(x_{1})=f(x_{2})$ (g is one to one - injective).

a) So, we see that $g(f(x_{1}))=g(f(x_{2}))$ means that  $x_{1}=x_{2}$, which means that $g\circ f$ is       an injective (one to one) map of A onto C.

Second, show the composite is surjective (onto):
by 1, we know that for any $b\in B$ there exists at least one $x\in A$ such that $f(x)=b$.
by 2, we know that for any $c\in C$ there exists at least one $b\in B$ such that $g(b)=c$.

b) So, we see that for any$c\in C$ there exists at least one  $x\in A$ such that $g(f(x))=c$, which means that    $g\circ f$ is a surjective (onto) map of A onto C.

With the results a) and b) we see that the composite $g\circ f$ is indeed a bijective map of A onto C.

QED