polarization
My guess:
Because polarization happens only when amplitude exists and because the sun is so big that the light rays arriving on earth literally come in all direction, their amplitude is then cancelled out.
…….but this doesn't make any sense, no amplitude means no light at all. No brightness. The earth would then be in darkness.
First let's deal with a false assumption:
similar to the way that the sum of a huge number of randomly selected 1's and -1's would never stray far from 0.
Suppose we have a set of $N$ random variables $X_i$, each independent and with equal probability of being either $+1$ or $-1$. Define $$ S = \sum_{i=1}^N X_i. $$ Then, yes, the expectation of $S$ may be $0$, $$ \langle S \rangle = \sum_{i=1}^N \langle X_i \rangle = \sum_{i=1}^N \left(\frac{1}{2}(+1) + \frac{1}{2}(-1)\right) = 0, $$ but the fluctuations can be significant. Since we can write $$ S^2 = \sum_{i=1}^N X_i^2 + 2 \sum_{i=1}^N \sum_{j=i+1}^N X_i X_j, $$ then more manipulation of expectation values (remember, they always distribute over sums; also the expectation of a product is the product of the expectations if and only if the factors are independent, which is the case for us for $i \neq j$) yields $$ \langle S^2 \rangle = \sum_{i=1}^N \langle X_i^2 \rangle + 2 \sum_{i=1}^N \sum_{j=i+1}^N \langle X_i X_j \rangle = \sum_{i=1}^N \left(\frac{1}{2}(+1)^2 + \frac{1}{2}(-1)^2\right) + 2 \sum_{i=1}^N \sum_{j=i+1}^N (0) (0) = N. $$ The standard deviation will be $$ \sigma_S = \left(\langle S^2 \rangle - \langle S \rangle^2\right)^{1/2} = \sqrt{N}. $$ This can be arbitrarily large. Another way of looking at this is that the more coins you flip, the less likely you are to be within a fixed range of breaking even.
Now let's apply this to the slightly more advanced case of independent phases of photons. Suppose we have $N$ independent photons with phases $\phi_i$ uniformly distributed on $(0, 2\pi)$. For simplicity I will assume all the photons have the same amplitude, set to unity. Then the electric field will have strength $$ E = \sum_{i=1}^N \mathrm{e}^{\mathrm{i}\phi_i}. $$ Sure enough, the average electric field will be $0$: $$ \langle E \rangle = \sum_{i=1}^N \langle \mathrm{e}^{\mathrm{i}\phi_i} \rangle = \sum_{i=1}^N \frac{1}{2\pi} \int_0^{2\pi} \mathrm{e}^{\mathrm{i}\phi}\ \mathrm{d}\phi = \sum_{i=1}^N 0 = 0. $$ However, you see images not in electric field strength but in intensity, which is the square-magnitude of this: $$ I = \lvert E \rvert^2 = \sum_{i=1}^N \mathrm{e}^{\mathrm{i}\phi_i} \mathrm{e}^{-\mathrm{i}\phi_i} + \sum_{i=1}^N \sum_{j=i+1}^N \left(\mathrm{e}^{\mathrm{i}\phi_i} \mathrm{e}^{-\mathrm{i}\phi_j} + \mathrm{e}^{-\mathrm{i}\phi_i} \mathrm{e}^{\mathrm{i}\phi_j}\right) = N + 2 \sum_{i=1}^N \sum_{j=i+1}^N \cos(\phi_i-\phi_j). $$ Paralleling the computation above, we have $$ \langle I \rangle = \langle N \rangle + 2 \sum_{i=1}^N \sum_{j=i+1}^N \frac{1}{(2\pi)^2} \int_0^{2\pi}\!\!\int_0^{2\pi} \cos(\phi-\phi')\ \mathrm{d}\phi\ \mathrm{d}\phi' = N + 0 = N. $$ The more photons there are, the greater the intensity, even though there will be more cancellations.
So what does this mean physically? The Sun is an incoherent source, meaning the photons coming from its surface really are independent in phase, so the above calculations are appropriate. This is in contrast to a laser, where the phases have a very tight relation to one another (they are all the same).
Your eye (or rather each receptor in your eye) has an extended volume over which it is sensitive to light, and it integrates whatever fluctuations occur over an extended time (which you know to be longer than, say, $1/60$ of a second, given that most people don't notice faster refresh rates on monitors). In this volume over this time, there will be some average number of photons. Even if the volume is small enough such that all opposite-phase photons will cancel (obviously two spatially separated photons won't cancel no matter their phases), the intensity of the photon field is expected to be nonzero.
In fact, we can put some numbers to this. Take a typical cone in your eye to have a diameter of $2\ \mathrm{µm}$, as per Wikipedia. About $10\%$ of the Sun's $1400\ \mathrm{W/m^2}$ flux is in the $500\text{–}600\ \mathrm{nm}$ range, where the typical photon energy is $3.6\times10^{-19}\ \mathrm{J}$. Neglecting the effects of focusing among other things, the number of photons in play in a single receptor is something like $$ N \approx \frac{\pi (1\ \mathrm{µm})^2 (140\ \mathrm{W/m^2}) (0.02\ \mathrm{s})}{3.6\times10^{-19}\ \mathrm{J}} \approx 2\times10^7. $$ The fractional change in intensity from "frame to frame" or "pixel to pixel" in your vision would be something like $1/\sqrt{N} \approx 0.02\%$. Even give or take a few orders of magnitude, you can see that the Sun should shine steadily and uniformly.
What you are seeing is stress in the window resulting in birefringence: the speed of propagation of polarized light depends on the direction of polarization.
In the setup you have, the light in the sky is partially polarized because that's how Rayleigh scattering works; this partially polarized light is transmitted through the window where it rotates (because of the birefringence) depending on the stress. The polarizer on your camera acts as the analyzer: some of the polarized light will be more at right angles while other light is more parallel to the axis of the second filter.
Now birefringence is a function of wavelength: so different colors will be rotated by different amounts, and will be more or less attenuated. And this is what gives rise to the colors.
Here is an example of an image of a plastic box with in built stresses viewed through crossed polarizers source:
UPDATE - why Rayleigh scattering leads to polarized light:
In this website we read:
The most common example of Rayleigh scattering is the scattering of visible radiation from the Sun by neutral atoms (mostly Nitrogen and Oxygen) in the upper atmosphere. The frequency of visible radiation is much less than the typical emission frequencies of a Nitrogen or Oxygen atom (which lie in the ultra-violet band), so it is certainly the case that $\omega \ll \omega_0$. When the Sun is low in the sky, radiation from it has to traverse a comparatively long path through the atmosphere before reaching us. Under these circumstances, the scattering of direct solar light by neutral atoms in the atmosphere becomes noticeable
In Rayleigh scattering, the electrons around an atom are a driven simple harmonic oscillator: classically, you can think of it as a negative cloud that can move with respect to a positive center, and if you could displace it, it would vibrate around its equilibrium position with some frequency $\omega_0$. Now when you excite this cloud with a transverse electrical signal (EM wave like light) it will emit light mostly at right angles to the axis of excitation - in fact there's a $\left(\frac{\omega}{\omega_0}\right)^4\sin^2\theta$ term in the intensity distribution. This both tells us that the intensity of the scattered light drops quickly for longer wavelengths (the key is blue) and also that when the sun is to your right, the polarization of the sky will be in the up/down direction (perpendicular to the line from the sun to the point). This is explained in more detail at this website on polarization which is also the source of this animation that shows the direction of polarization that you expect:
Best Answer
The light that we receive directly from the Sun is emitted in the photosphere, travels in a straight line through the atmosphere to us, and is unpolarised. The photosphere is a region covering the whole surface of the Sun and is also of order hundreds of km thick. Thus the light received at a point on Earth is a pseudo-random mixture from many points on the Sun, each of which was emitted in an independent process. There is little to make these photons coherent in any way, and a random mixture of polarisations at random phases just gives unpolarised light.
That being said, there are regions of the Sun (namely sunspots) that do emit partially polarised light. This is caused by the strong magnetic fields and Zeeman splitting or broadening of light at wavelengths corresponding to atomic transitions. The split components have their own peculiar circular or plane polarisations This effect is used to estimate the overall magnetic flux from the surfaces of very magnetically active stars or even to map their magnetic fields through a technique called Zeeman Doppler Imaging.
If your question extends to considering sunlight scattered by the atmosphere - i.e. the sky - then the light is partially polarised. Unpolarised light entering the atmosphere can be considered an equal mix of two perpendicular plane polarisation states. Rayleigh scattering through 90 degrees would select one of these polarisation states, causing us to see almost perfectly polarised light from that direction. Smaller scattering angles cause less polarisation, whilst scattering from a variety of heights in the atmosphere means that the polarisation is never perfect.