# Cohomology and exterior algebra

Why I can look at $H^{*}(S^{3})$ as the exterior algebra $\Lambda(x_{3})$? Where $x_{3} \in H^{3}(S^{3})$ is a cohomology class (I suppose...).

-
What makes you think you can? –  Michael Albanese Feb 3 '13 at 10:48
@MichaelAlbanese I red that $SU(2) \simeq S^{3}$ (and I am agree), but I don't understand why $H^{*}(SU(2)) \simeq \Lambda(x_{3})$, where $x_{3} \in H^{3}(S^{3})$... –  ArthurStuart Feb 3 '13 at 11:21
@MichaelAlbanese google.com.mx/url?q=http://www-irma.u-strasbg.fr/~oancea/… Page 4 first row... –  ArthurStuart Feb 3 '13 at 11:41

I assume you are working over a fixed field $k$, to simplify the matters.
By the universal property of exterior algebra (see wikipedia) there is a unique homomorphism of algebras $\tilde{i}: \Lambda(x_{3}) \rightarrow H^{*}(S^{3})$ induced by the inclusion $i: span(x_{3}) \rightarrow H^{*}(S^{3})$, because $x_{3} ^{2} = 0$ in $H^{*}(S^{3})$.
This is an isomorphism by dimension count. Since $dim _{k}(\Lambda(x _{3})) = dim _{k}(H^{*}(S^{3})) = 2$, it is enough to show that $\tilde{i}$ is surjective. Notice that $1 \in H^{0}(S^{3})$ is in the image of $\tilde{i}$, because it is an algebra homomorphism and $x_{3} \in H^{3}(S^{3})$ is in the image by construction. These together span $H^{*}(S^{3})$, ending the argument. (This works for all positive-dimensional spheres, there is nothing special about $S^{3}$).
So can I use that to calculate the cohomology of the bundle $U(n-1) \hookrightarrow U(n) \rightarrow S^{2n-1}$? –  ArthurStuart Feb 3 '13 at 12:43
Pstagowski But what is in this example $x_{3}$ and $\Lambda(x_{3})$? And why $x_{3}^{2}=0$? –  ArthurStuart Feb 3 '13 at 13:58
I used your notation, so $x_3$ is a non-zero cohomology class in $H^{3}(S^{3})$. It you are working over a field, it must generate it, so $\Lambda(x_{3}) = \Lambda(span(x_{3})) = \Lambda(H^{3}(S^{3}))$. We have $x_{3}^{2} = 0$, because it has degree $6$, but $H^{6}(S^{3}) = 0$. About the computation of the cohomology of the bundle - I guess you can work that out, but I know of no easy way to compute cohomology of a fibre bundle. You can google for Serre spectral sequence. –  Piotr Pstrągowski Feb 4 '13 at 11:08