I need help solving something I don't understand. OK so the problem is this:
$$H^k(X,C)=\bigoplus_{p+q=k} H^{p,q}(X),$$
What does the $\;p+q=k\;$ mean?
Thank you anybody that helps! :)
I need help solving something I don't understand. OK so the problem is this:
$$H^k(X,C)=\bigoplus_{p+q=k} H^{p,q}(X),$$
What does the $\;p+q=k\;$ mean?
Thank you anybody that helps! :)
On
There are multiple ways to indicate a summation (similarly with a product, direct summation, etc.) In all cases, the sum is evaluated in all ways such that the condition is met, subject to natural restrictions that are implicit in the problem. Typical restrictions are that the index is an integer, and sometimes that the index is nonnegative.
Examples: $$\sum_{i\ge 0} f(i)$$ and $$\sum_{i\in[0,6]}f(i)$$ and $$\sum_{j|n} f(j)$$ and $$\sum_{j|n, j\neq 3} f(j)$$ and $$\sum_{jk=n} f(j)f(k)$$ and $$\sum_{j+k=n} f(j)$$
On
There is something to be said for writing $$ \bigoplus_{p,\,q\, :\, p+q=k} H^{p,q}(X). $$ For example, in case $k=3$, this would mean $$ H^{0,3}(X) \oplus H^{1,2}(X)\oplus H^{2,1}(X)\oplus H^{3,0}(X) $$ (or if only positive indices were indicated by the context, then discard the first and last terms, etc.).
Sometimes people write $$ \sum_{i<j} a_{ij} $$ when $j$ is fixed by the context and it means $i$ runs through the list of all values less than $j$. But sometimes they mean the pair $i,j$ runs through the list of all values in which $i<j$, and neither is fixed. It is sometimes convenient to make the notation explicit about the difference, writing $$ \sum_{i\,:\,i<j} a_{ij} $$ in one case, and $$ \sum_{i\,j\,:\,i<j} a_{ij} $$ in the other.
$p+q = k$ indicates that the sum of indexes $p, q$ is equal to $k$, where $k$ is constant, for the given $k$. If $k=4$ you can take $p=1, q=3$ or $p = q = 2,$ etc.
So every possible pair $(p, q)$ for which $p+q = k$; where $p, q$ are the indices which indicate which $H^{p, q}(X)$ to sum.
From comment
Indeed: to "spell it out the notation" you've posted, for a particular value of $k$, let's let $k = 4.\;$ Then: $$H^4(X,C)=\bigoplus_{p+q=k} H^{p,q}(X) = H^{0, 4}(X) \oplus H^{1, 3}(X) \oplus H^{2,2}(X) \oplus H^{3,1}(X)\oplus H^{4,0}(X)$$