Sums, Quotients, and Complexes

Sums and Quotients

Let V and W be vector spaces over a field \mathbf{F}.

The direct sum of V and W is the set of pairs V\oplus W = \{(v,w) : v\in V, w\in W \}

This is a vector space with the operations (v,w) + (v',w') = (v+v', w+w').

and

a(v,w) = (av,aw).

One can define the direct sum of a finite family \{V_{i}\}_{i=1}^{m} of vector spaces to be the set of sequences

(v_1,\ldots, v_m)

where v_{i}\in V_{i}.

Again the operations are componentwise.

Infinite direct sums are also defined. Given an index set I and a family \{V_{i}\}_{i\in I} of vector spaces, then the direct sum \oplus_{i\in I}V_{i}

is the set of functions f:I\to \bigcup_{i\in I}V_{i} such that f(i)\in V_{i} and f(i)=0 for all but finitely many i\in I.

The field \mathbf{F} is a one-dimensional vector space over itself. The vector space \mathbf{F}^{n} is just \oplus_{i=1}^{n}\mathbf{F}.

Note that V and W are subspaces of V\oplus W by identifying V with the pairs (v,0) and W with the pairs (0,w). The intersection of these two subspaces is 0=(0,0).

Proposition: If V and W are finite dimensional of dimension n and m respectively, then V\oplus W is finite dimensional of dimension n+m.

In fact if v_1,\ldots, v_n is a basis for V, and w_1,\ldots, w_m is a basis for W, then (v_1,0),\ldots, (v_n,0),(0,w_1),\ldots, (0,w_m) is a basis for V\oplus W.

They span since (v,w) = (\sum a_{i}v_{i}, \sum b_{i}w_{i}) = \sum a_{i}(v_{i},0) + \sum b_{i}(0,w_{i}).

They are independent since if (\sum a_{i}v_{i},\sum b_{i}w_{i})=(0,0)

then all the a_{i} and b_{i} are zero by independence of the v’s and w’s.

Subspace sums and internal direct sums

Let V be a vector space and let U and W be subspaces of V. The sum of U and W is the set U+W = \{u+w : u\in U, w\in W\}

This is a subspace of V.

It is the smallest subspace of V containing both W and U.
If u_1,\ldots, u_n span U and w_1,\ldots, w_k span W then the union u_1,\ldots, u_n, w_1,\ldots, w_k span U+W.
the dimension of U+W is at most \mathrm{dim}(U)+\mathrm{dim}(W). It can certainly be smaller (suppose U=W for example).

Algorithm to compute a basis for U+W.

Choose a basis for V and take spanning sets as above for U and W. Create a matrix A whose columns are the u_i and w_j written in the chosen basis for V.
Put the matrix A in rref yielding A'.
The columns of A corresponding to the pivot columns of A' are a basis for U+W.
The basis constructed in this way will be a subset of the union of the u_{i} and w_{j}.

Internal direct sum

If U and W are subsets of V such that every element of x\in U+W has a unique representation x=u+w with u\in U and w\in W, then U+W is an (internal) direct sum and U+W is isomorphic to U\oplus W.

Proposition: U+W is a direct sum if and only if either of the following hold:

U\cap W=0.
\mathrm{dim}(U)+\mathrm{dim}(W)=\mathrm{dim}(U+W).

Suppose that u+w=u'+w' for u,u'\in U and w,w'\in W. Then (u-u') = (w'-w). The left side is in U, and the right side is in W, so both u-u' and w'-w are in U\cap W. If U\cap W=0 this implies that u=u' and w=w' so the representation is unique. If not, then choose z\in U\cap W with z\not=0 and we have u+w = (u+z) + (w-z) so the representation is not unique.

For the second, choose bases u_1,\ldots, u_n of U and w_1,\ldots, w_k of W. Together these span U+W. On the other hand if there is a non trivial linear relation among them, we would have \sum a_{i}u_{i} +\sum b_{j}w_{j}=0. Since the u_i and w_j are independent, any such relation must involve a mix of both u_i and w_j vectors, and so represents an element of U\cap W.

So the formula above is just the rank plus nullity = columns equation for the matrix A. At least one a_{i} and one b_{j} must be nonzero (since the u’s and w’s are separately independent). This gives us a non-zero vector \sum a_{i}u_{i} = -\sum b_{j}w_{j} in U\cap W.

Thus they are independent and (b) follows. Conversely, if we have the relation on dimensions, these u_i and w_{j} must be independent. A nonzero element of U\cap W would give a non-trivial equation \sum a_{i}u_{i} = \sum b_{j} w_{j}

which would imply the u_i and v_j are dependent and thus the dimension of U+W is smaller than n+k.

One can improve this result a bit to obtain this formula.

Proposition: Given subspaces U and W of V we have \begin{aligned} \mathrm{dim}(U+W) &= \mathrm{dim}(U\oplus W)-\mathrm{dim}(U\cap W) \\ &= \mathrm{dim}(U)+\mathrm{dim}(W)-\mathrm{dim}(U\cap W) \end{aligned}

This follows from our algorithm above. Make a matrix A whose columns are the basis elements u_i and w_j for U and W respectively written in some basis for V. The rank of A is dimension of the span of the columns of A, which is the dimension of U+W. Let y^{(k)} be a basis for the null space of A. Each element of this basis gives a relation \sum_{i=1}^{\mathrm{dim}(U)} y^{(k)}_{i}u_{i} + \sum_{j=1}^{\mathrm{dim}(W)}y^{(k)}_{\mathrm{dim}(U)+j}w_{j}=0. We can read this relation as: \sum_{i=1}^{\mathrm{dim}(U)} y^{(k)}_{i}u_{i} = - \sum_{j=1}^{\mathrm{dim}(W)}y^{(k)}_{\mathrm{dim}(U)+j}w_{j}. Since the u_{i} and w_{j} are linearly independent, it must be the case that at least one of the y_{i} on the left, and one on the right, are non-zero. Therefore each element y^{(k)} of the null space of A yields an h_{k}=\sum_{i=1}^{\mathrm{dim}(U)} y^{(k)}_{i}u_{i} that is a non-zero element that belongs to U\cap W.

We claim that the h_{k} are a basis for U\cap W. The spanning property comes because if h\in U\cap W, then writing h=\sum a_{i}u_{i}=\sum b_{j}w_{j} in terms of the u_{i} and in terms of the w_{j} gives a relation \sum a_{i}u_{i}-\sum b_{j}w_{j}=0 which in turn produces an element of the null space of A. Writing this in terms of the y^{(k)} in turn yields a linear expression for h in terms of the h_{k}.

To see the independence of the h_{k}, suppose \sum_{k} a_{k}h_{k} = \sum_{k}\sum_{i} a_{k}y^{(k)}_{i}u_{i} = 0. The independence of the u_{i} means that, for each i=1,\ldots,\mathrm{dim}(U), \sum_{k} a_{k}y^{(k)}_{i}=0. A similar expression using the w_{j} gives \sum_{k} a_{k}y^{(k)}_{\mathrm{dim}(U)+j}=0 for all j=1,\ldots, \mathrm{dim}(W). Putting these together gives a relation among the y^{(k)}, which are independent, so all a_{k} must be zero.

We’ve shown that the rank of A is \mathrm{dim}(U+W), the number of columns of A is \mathrm{dim}(U)+\mathrm{dim}(W), and the nullity of A is \mathrm{dim}(U\cap W), so the desired formula is the known formula \mathrm{rank}(A)+\mathrm{nullity}(A) = \mathrm{columns}(A)

\begin{aligned} \mathrm{dim}(U+W) &= \mathrm{dim}(U\oplus W)-\mathrm{dim}(U\cap W) \\ &= \mathrm{dim}(U)+\mathrm{dim}(W)-\mathrm{dim}(U\cap W) \end{aligned}

Quotient Spaces

An equivalence relation on a set X is a relation x\sim y between any two elements of X with the properties that:

x\sim x for all x\in X.
If x\sim y, then y\sim x for all x,y\in X.
If x\sim y and y\sim z then x\sim z for all x,y,z\in X.

An equivalence relation partitions X into equivalence classes. An equivalence class [x] =\{y\in X : y\sim x\}

Every element of X belongs to exactly one class. Put another way, given two elements x and y in X, either [x]=[y] or [x]\cap [y]=\emptyset.

To see this, clearly x\in [x]. Suppose a\in [x]\cap [y]. Then a\sim x and a\sim y so y\sim x. Therefore y\in [x]. If u\sim y, then u\sim x so u\in [x]. Therefore [y]\subset [x]. The same argument run “the other way” shows [x]\subset [y]. So [x]\cap [y]\not=\emptyset implies [x]=[y].

Now suppose V is a vector space over F and W is a subspace of V. Define an equivalence relation on V by setting x\sim y \iff x-y\in W.

The set of equivalence classes [x] is called the quotient space V/W.

In fact V/W is a vector space over F. We define [x]+[y] = [x+y] and a[x]=[ax]

We have to verify that these operations are well-defined and that they satisfy all the vector space axioms.

To see that addition is well-defined, suppose [x]=[a] and [y]=[b]. Then we claim [x+y]=[a+b] This is because (a+b)-(x+y)=(a-x)+(b-y)\in W.

Similarly if [x]=[y] then x-y\in W so ax-ay\in W and therefore [ax]=[ay].

Checking all the axioms is tedious but straightforward. For example, \begin{aligned} a([x]+[y])&=a([x+y])=[a(x+y)] \\ &=[ax+ay]=[ax]+[ay]=a[x]+a[y] \end{aligned}

Proposition: Suppose that V is finite dimensional and W\subset V. Then \mathrm{dim}(V/W)=\mathrm{dim}(V)-\mathrm{dim}(W).

Proof: The map f:V\to V/W defined by f(v)=[v] is an onto linear map with nullspace exactly W. Linearity:

f(x+y)=[x+y]=[x]+[y]=f(x)+f(y)
f(ax) = [ax] = a[x] = af(x)

The fact that the map is onto is because if [v]\in V/W then f(v)=[v].

Finally, the nullspace of f is the set of v such that f(v)=[v]=[0]. But [v]=[0] if and only if v\in W.

Thus the dimension relation follows from the rank/nullity theorem for the linear map f.

Proposition: Let f:V\to W be a linear map and suppose H\subset V is contained in the kernel of f. Then there is a unique linear map \overline{f}:V/H\to W such that \overline{f}([v])=f(v) for all v\in V.

Proof: The map \overline{f} defined by \overline{f}([v])=f(v) has the desired property and is well-defined because if [v]=[x] then v-x\in H. Since H is contained in the kernel of f, this means that f(v-x)=0 so f(v)=f(x).

The existence of the map \overline{f} is called the universal mapping property of the quotient V/H.

One can show that the subspaces of V/H correspond exactly to the subspaces of V containing H.

Complexes

Definition: A complex of vector spaces is a sequence of vector spaces W^{i} connected by linear maps d_{i} \cdots \to \cdots \to W^{i}\stackrel{\to}{d_{i}} W^{i+1} \stackrel{\to}{d_{i+1}}W^{i+2}\to\cdots

with the property that d_{i+1}d_{i}=0 for all i.

The complex is bounded if all but finitely many W^{i} are zero.

Let Z^{i}\subset W^{i} be the null space of d_{i} and let B^{i}\subset W^{i} be the image d_{i-1}(W_{i-1})\subset W^{i}.

The elements of Z^{i} are called cocycles and those of B^{i} are called coboundaries.

The vector space Z^{i}/B_{i} is the i^{th} cohomology H^{i}(W^{\cdot}) of the complex.

Suppose the complex is bounded and all W^{i} are finite dimensional. The sum \chi(W^{\cdot})=\sum_{-\infty}^{\infty} (-1)^{i}\mathrm{dim}(W^{i})

is called the Euler characteristic of the complex.

Proposition: We have \chi(W^{\cdot})=\sum_{-\infty}^{\infty} (-1)^{i}\mathrm{dim}H^{i}(W^{\cdot})

Proof: Notice that \mathrm{dim}(B^{i+1})=\mathrm{dim}(W^{i})-\mathrm{dim}(Z^{i}) since the image of d^{i} is B^{i+1} and its kernel is Z^{i}. So: \sum (-1)^{i}\mathrm{dim}(W^{i}) = \sum (-1)^{i}(\mathrm{dim}(B^{i+1})+\mathrm{dim}(Z^{i})). Now \sum (-1)^{i}(\mathrm{dim}(B^{i+1})+\mathrm{dim}(Z^{i}))=\sum (-1)^{i}(\mathrm{dim}(Z^{i})-\mathrm{dim}(B^{i}))

which is the alternating sum of the dimensions of the H^{i}(W^{\cdot}).

Definition: A complex is exact if its cohomology groups are zero. In this case, the kernel of d^{i} is exactly the image of $d^{i-1}.

So the cohomology of a complex is a measure of how far it is from being exact.

A short exact sequence is a 5 term exact complex: 0\to A\to B\to C \to 0.

In this case, the map from A\to B is injective (its kernel is zero) and the map from B\to C is surjective.

From the universal property of quotients, this means that C is isomorphic to A/B.

Homology

One can also define a complex with the reverse indexing, so that d_{i}:W_{i}\to W_{i-1}. In this case the kernel Z_{i} of d_{i} is called the space of i-cycles, and the image of d_{i+1} is called the space of i-boundaries. The quotient Z_{i}/B_{i} is called the i^{th} homology group of the complex H_{i}(W^{\cdot}).

Maps of Complexes

A linear map f between complexes \{W^{\cdot}\} and \{V^{\cdot}\} is a family of linear maps f_{i}: W^{i}\to V^{i} with the property that f_{i+1}\circ d_{i}=d_{i}\circ f_{i}.

If f is such a linear map, then it carries cocycles to cocycles and coboundaries to coboundaries. For coboundaries, this is because if a_{i+1}=d_{i}(a_{i}) then f_{i+1}(a_{i+1}) = f_{i+1}d_{i}(a_{i})=d_{i+1}(f_{i}(a_{i})) For cocycles, this is because, if d_{i}(a_{i})=0, then d_{i}(f_{i}(a_{i})) then d_{i}(f_{i}(a_{i})=f_{i+1}(d_{i}(a_i)) = 0.

Thus from f we obtain a family of maps \overline{f}:H^{i}(W^{\cdot})\to H^{i}(V^{\cdot}).

This is called the induced map on cohomology.