I am filled with all kinds of vector space and I want to make sure I understand the basis for each kind of vector space.

Suppose $\{v_i\}_{i=1}^n$ is the basis for vector space $V$, $\{w_j\}_{j=1}^m$ is the basis for vector space $W$.

  1. $V+W$: all linear combo of $\{v_i\}$, $\{w_j\}$, basis is $\{v_i,w_j\}$, dimension is $\dim(V)+\dim(W)-\text{number of overlap}$
  2. $V\oplus W$: Same as $V\times W$?
  3. $V\times W$: all $(v,w)$ with $v\in V$ and $w\in W$, basis $(v_i,w_j)$, dimension is $\dim(V)\dim(W)$
  4. $V\otimes W$: all linear combo of $v_i\otimes w_j$, basis $v_i\otimes w_j$, dimension is $\dim(V)\dim(W)$. By universal property of tensor, $V\times W$ is isomorphism to $V\otimes W$?
  5. $V\bigwedge V$ or $\bigwedge^rV$: all linear combo of $v_i\wedge v_j$, basis $v_i\wedge v_j$ and $i\neq j$, dimension is ${\dim V}\choose r$
  6. $V^{`}\cdot V$ or $S_rV$:(symmetric power) all linear combo of $v_i^{`}\cdot v_j$, basis $v_i^{`}\cdot v_j$ and $i\neq j$, dimension is ${\dim V +r-1} \choose r$.

So eventually, I want to understand what are the basis for the two vector spaces $\bigwedge^2(V\oplus W)$ and $ \bigwedge^2V\oplus(V\otimes W)\oplus\bigwedge^2W$(and finally I need to find an isomorphism between these two spaces, which need to prove it maps basis to basis, and that is why I need to make sure I know the basis).

Let $\{(v_i,w_j)\}=\{e_k\}$, where $i=1,\cdots,n$, $j=1,\cdots,m$, $k=1,\cdots,nm$.

Is $\{e_i\wedge e_j\}_{i,j=1}^{nm}$ with $i\neq j$ the basis for $\bigwedge^2(V\oplus W)$? OMG, I am so confused.

BTW, it seems that these two spaces are not isomorphism, is the problem wrong?

  • 1
    $\begingroup$ In (1), $V$ and $W$ need not be complimentary; consider the case $V + V$. In (3), $\dim (V \times W) = \dim V + \dim W$. In general, $V \times W$ is not isomorphic to $V \otimes W$. In (5), $\Lambda^r$ has basis elements $v_{i_1} \wedge \cdots \wedge v_{i_r}$, $i_1 < \cdots < i_r$. There's a similar issue in (6) with symmetric powers with $r > 2$. $\endgroup$ – Travis Willse Apr 7 '15 at 13:44
  • $\begingroup$ @DietrichBurde Well, we can assume that, since my actual problem is just about finite dimension. $\endgroup$ – breezeintopl Apr 7 '15 at 13:59
  • 2
    $\begingroup$ All right. If you say "Suppose $\{v_i \}$ is the basis", then infinite dimension is allowed. You can add $i=1,\ldots ,n$. $\endgroup$ – Dietrich Burde Apr 7 '15 at 14:02
  • $\begingroup$ @Travis Thank you for your answer! I fixed (1). For (3), why it is the sum not the product? I think for $(v_i,w_j)$, the first element has $dim(V)$ choices, and the second element has $dim(W)$ choices, so it should be $dim(V)dim(W)$. Am I wrong? BTW, just want to make sure, $V\oplus W$ and $V\times W$ are the same thing, right? $\endgroup$ – breezeintopl Apr 7 '15 at 14:04
  • $\begingroup$ @breezeintopl Yes, $V \oplus W \cong V \times W$, see math.stackexchange.com/questions/39895/… for more. In both cases, given a basis $B$ of $V$ and a basis $C$ of $W$, the union $\{(b, 0) : b \in B\} \cup \{(0, c) : c \in C\}$ is a basis of the sum/product. NB this gives $\dim(V \oplus W) = \dim V + \dim W$ as desired. $\endgroup$ – Travis Willse Apr 7 '15 at 14:28

As in the question, we set $n := \dim V$ and $m := \dim W$ and assume both are finite (though many conclusions hold just as well when either or both are infinite), and pick arbitrary bases $A = (v_a)$ of $V$ and $B = (w_b)$ of $W$. Also, we usually regard bases as ordered sets of vectors; the below doesn't always specify orders, but of course these can be chosen arbitrarily to produce a bona fide basis.

  1. Sums $V + W$ Pick a basis $(x_c)$ of $V \cap W$ (denote $p := \dim (V \cap W) = p$), and extend it to a basis $(x_1, \ldots, x_p, y_1, \ldots, y_{m - p})$ of $W$. Then, $$(v_1, \ldots, v_n, y_1, \ldots, y_{m - p})$$ is a basis of $V + W$, which thus has dimension $n + m - p$; usually this is written $$\dim(V + W) = \dim V + \dim W - \dim (V \cap W).$$

  2. Direct sums $V \oplus W$ Via some identifications we can view this as a special case of (1), but we may as well be explicit: $$(A \times \{0\}) \cup (\{0\} \times B) = \{(v_a, 0) : v_a \in A\} \cup \{(0, w_b) : w_b \in B\}$$ is a basis of $V \oplus W$, and so $$\dim (V \oplus W) = \dim V + \dim W.$$ For convenience, we often mildly abusively write $(v_a, 0)$ as $v_a$ and $(0, w_a)$ as $w_a$, in which case we can write the basis as $$A \cup B = (v_1, \ldots, v_n, w_1, \ldots, w_m).$$

  3. Cartesian products $V \times W$ Since the product has a finite number of factors, this is indeed the same as $V \oplus W$ in (2); see The direct sum $\oplus$ versus the cartesian product $\times$ for more.

  4. Tensor products $V \otimes W$ $$\{v_a \otimes w_b : v_a \in A, w_b \in B\} \cong A \times B$$ is a basis for $V \otimes W$ and hence $$\dim (V \otimes W) = \dim V \cdot \dim W.$$

  5. Multivector spaces $\bigwedge^k V$ $$\{v_{a_1} \wedge \cdots \wedge v_{a_k} : 1 \leq a_1 < \cdots < a_k \leq n\},$$ and so $$\dim \bigwedge^k V = {{\dim V}\choose{k}}.$$ By convention, $\bigwedge^0 V$ is just the field $\mathbb{F}$ underlying $V$, and we take the empty wedge product to be $1 \in \mathbb{F}$, so $(1)$ is a basis of $\bigwedge^0 V$; $\bigwedge^k V$ is trivial for $k > m$, in which case the given basis is empty.

  6. Symmetric powers $\bigodot^k V$ $$\{v_{a_1} \odot \cdots \odot v_{a_k} : 1 \leq a_1 \leq \cdots \leq a_k \leq n\}$$ is a basis for $\bigodot^k V$, where $\odot$ defines the symmetric product, and one can then show $$\dim \bigodot^k V = {{\dim V + k - 1}\choose{k}}.$$

For completeness we should also mention one more:

Tensor powers $\bigotimes^k V$ $$\{v_{a_1} \otimes \cdots \otimes v_{a_k} : v_{a_1}, \ldots, v_{a_k} \in A\} \cong A \times \cdots \times A$$ is a basis for $\bigotimes^k V$, so that $$\dim \bigotimes^k V = (\dim V)^k.$$

Bivector space over a direct sum Finally, we can treat $\bigwedge^2 (V \oplus W)$. By (2), a basis for $V \oplus W$ is $$A \cup B = \{v_1, \ldots, v_n, w_1, \ldots, w_m\},$$ and then by (5), a basis of $\bigwedge^2 (V \oplus W)$ is \begin{multline} (v_1 \wedge v_2, \ldots, v_1 \wedge v_n, v_1 \wedge w_1, \ldots, v_1 \wedge w_m, v_2 \wedge v_3, \ldots, w_{m - 1} \wedge w_m) \\ = \{v_a \wedge v_{a'} : a < a'\} \cup \{v_a \wedge w_b\} \cup \{w_b \wedge w_{b'} : b < b'\}. \end{multline}

The elements $v_a \wedge v_{a'}$ comprise a basis for $\bigwedge^2 V$, and likewise those of the form $w_b \wedge w_{b'}$ comprise a basis for $\bigwedge^2 W$. The forms $v_a \wedge w_b$ span a space that can be identified with $V \otimes W$ via the map characterized by the isomorphism $v_a \otimes w_b \mapsto v_a \wedge w_b$. This gives a natural decomposition $$\bigwedge^2 (V \oplus W) \cong \bigwedge^2 V \oplus (V \otimes W) \oplus \bigwedge^2 W.$$ Of course, the dimension formulas produced above lead to the same result for the dimension of $\bigwedge^2 (V \oplus W)$: $${{n + m}\choose 2} = \frac{1}{2}(n + m)(n + m - 1) = \frac{1}{2}n(n - 1) + nm + \frac{1}{2}m(m - 1) = {n \choose 2} + nm + {m \choose 2}.$$

  • $\begingroup$ What is $r$ is in $(6)$ (symmetric powers) ? Is it $k$ ? Also what is $(y_c)$ ? $\endgroup$ – Why Sep 20 '20 at 14:44
  • 1
    $\begingroup$ @MabudAli Yes, $r$ should be $k$ (now corrected). I've simplified the statement in part (1) and in particular removed explicit reference to $(y_c)$. $\endgroup$ – Travis Willse Sep 21 '20 at 6:12

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