I'm trying to understand how to decompose certain symmetric and anti-symmetric tensor products of vector spaces into direct summands.

Let $V$ be a complex finite dimensional vector space and denote by $\Lambda^2 V$ and $S^2 V$ the spaces generated by $\frac{1}{2}(v_1 \otimes v_2 - v_1 \otimes v_2)$ and $\frac{1}{2}(v_1 \otimes v_2 + v_1 \otimes v_2)$ respectively (for all $v_1,v_2 \in V$).

At the moment I'm just trying to understand one concrete example. If we take $V = \mathbb{C}^5$ and we decompose $V = \mathbb{C}^3 + \mathbb{C}^2$ my guess is that the tensor product decomposes as:

$$\Lambda^2 V = \Lambda^2 (\mathbb{C}^3 \oplus \mathbb{C}^2) = \Lambda^2\mathbb{C}^3 \oplus \Lambda^2\mathbb{C}^2 \oplus (\mathbb{C}^3 \oplus \mathbb{C}^2)$$

I've arrived at that by thinking of the $\Lambda^2 V$ as an anti-symmetric matrix and then decomposing it blockwise. I'm struggling to prove the above statement in general and I'm trying to understand how $\Lambda^2$ and $S^2$ behave with direct sum decompositions generally.



I am guessing that your question is: What is $\bigwedge^k (V \oplus W)$? (This is the wedge product, not the tensor product.)

And also the same question for $S^k = Sym^k$, which is the symmetric product.

Extended hint:

The thing to observe is that there is a natural basis for $V \oplus W$. Namely, you take a basis for $V$ and union it with a basis for $W$.

Let's call our natural basis for $V$ $\{v_1, \ldots, v_n\}$, and our basis for $W$ $\{w_1, \ldots, w_m\}$.

Now, given a basis for a vector space $X$, there is a natural basis for $\bigwedge^k X$ and for $Sym^k X$:

Suppose that $\{x_1, \ldots, x_r\}$ is a basis for $X$.

1) Then a basis for $\bigwedge^k X$ is given by all of the $\{x_{i_1} \wedge \ldots \wedge x_{i_k} \}$ for all $i_1 < \ldots < i_k$, $i_j \in \{1, 2, \ldots r\}$.

2) A natural basis for $Sym$ is given similarly, only now you are allowed to take repeated vectors (so $<$ will be replaced by $\leq$) - another description is the set of monomials of degree k in $\mathbb{Z}[x_1, \ldots, x_r]$.

Finally: Given that $\{x_1, \ldots, x_r\} = \{v_1, \ldots, v_n, w_1, \ldots, w_m\}$ is a basis for $V \oplus W = X$, you can now play a combinatorial game to divide up $\bigwedge^k X$ into direct sums of smaller exterior powers of $V$ and $W$. Similarly for the symmetric product. Do you see how to proceed? Please feel free to ask if you have questions.

  • $\begingroup$ Thanks for your answer and clarifying my question! I'm struggling a little to finish this off - I can't really see who we can divide $\Lambda^k X$ or $S^k X$ into smaller exterior powers? $\endgroup$ – Wooster Jan 2 '16 at 17:06
  • $\begingroup$ @Wooster For example, consider basis given by selecting $a$ basis vectors from the $v_i$ piles, and $b$ basis vectors from the $w_j$ pile. $\endgroup$ – Lorenzo Jan 2 '16 at 17:07
  • $\begingroup$ Ah I see. So we have $\Lambda^k X = \bigoplus_{i+j=k, i \neq j} \Lambda^iV \otimes \Lambda^j W$ and $S^k X = \bigoplus_{i+j=k} \Lambda^iV \otimes \Lambda^j W$? $\endgroup$ – Wooster Jan 2 '16 at 17:21
  • $\begingroup$ @Wooster The first is close, but why disallow $i = j$? Just try to write out the basis. The second is also close, but $\wedge^i V$ gives anti-commuting variables in the polynomial ring analogy. Try to write out the basis elements explicitly (say in some simple case like $\mathbb{C}(e_1, e_2)$ and $\mathbb{C}(f_1, f_2)$ as your $V$ and $W$) and match them up. You can also compute the dimensions to sanity check. (Tensoring multiplies dimensions, direct sum adds them. What is $dim \wedge^k V$ and $dim Sym^k V$, given $dim V = n$? Hint: Look for a nice basis.) $\endgroup$ – Lorenzo Jan 3 '16 at 0:58
  • $\begingroup$ Thanks for your help! The dimension of $\Lambda^k V$ should be $d$ choose $k$ because we have to choose $k$ different basis elements from a $d$ dimensional space for each basis element - we can't have any repeats. I'm struggling to see how we can get the dimension of $S^kV$ actually? Playing around with those simple examples would suggest $\Lambda^kX = \bigoplus_{i+j=k} \Lambda^i V \otimes \Lambda^jW$ and the same for the symmetric case with $\Lambda$ replaced with $S$? $\endgroup$ – Wooster Jan 3 '16 at 9:15

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